Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C

Other Exercises

Question 1.
Find AD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q1.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q1.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q1.3

Question 2.
In the following diagram.
AB is a floor-board. PQRS is a cubical box with each edge = 1 m and ∠B = 60°. Calculate the length of the board AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q2.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q2.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q2.3

Question 3.
Calculate BC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q3.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q3.2

Question 4.
Calculate AB .
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q4.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q4.2

Question 5.
The radius of a circle is given as 15cm and chord AB subtends an angle of 131° at the centre C of the circle. Using trigonometry, Calculate :
(i) the length of AB;
(ii) the distance of AB from the centre C.
Solution:
Chord AD substends an angle of 131° at the centre. Join CA, CB and draw CD ⊥ AB which bisects AB at D.
(In ∆CAB)
∵ CA = CB (radii of the same circle)
∴ ∠CAB = ∠CBA
But ∠CAB + ∠CBA = 180°- 131° = 49°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q5.2

Question 6.
At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is \(\frac { 5 }{ 12 }\). On walking 192 metres towards the tower; the tangent of the angle is found to be \(\frac { 3 }{ 4 }\). Find the height of the tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q6.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q6.3

Question 7.
A vertical tower stands on horizontal plane and is surmounted by a vertical flagstaff of height h metre. At a point on the plane, the angle of elevation of the bottom of the flagstaff is a and that of the top of flagstaff is β. Prove that the height of the tower is :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q7.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q7.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q7.4

Question 8.
With reference to the given figure, a man stands on the ground at point A, which is on the same horizontal plane as B, the foot of the vertical pole BC. The height of the pole is 10 m.
The man’s eye is 2 m above the ground. He observes the angle of elevation of C. The top of the pole, as x°, where tan x° = \(\frac { 2 }{ 5 }\) . calculate:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q8.1
(i) the distance AB in m;
(ii) the angle of elevation of the top of the pole
when he is standing 15 m from the pole. Give your answer to the nearest degree. [1999]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q8.3

Question 9.
The angles of elevation of the top of a tower from two points on the ground at distances a and b metres from the base of the tower and in the same striaght line with it are complementary.Prove that height of the tower is \( \sqrt{ab} \) metre.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q9.2

Question 10.
From a window A. 10 m above the ground the angle of elevation of the top C of a tower is x°, where tan x = \(\frac { 5 }{ 2 }\) and the angle of depression of the foot D of the tower is y°, where tany° = \(\frac { 1 }{ 4 }\).(See the figure given below). Calculate the height CD of the tower in metres. [2000]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q10.1
Solution:
Let CD be the height of the tower and height of window A from the ground = 10m
In right ∆AEC,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q10.3

Question 11.
A vertical tower is 20 m high, A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower ? [2001]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q11.2

Question 12.
A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60°. When he moves 50 m away from the bank, he finds the angle of elevation to be 30°. Calculate :
(i) the width of the river and
(ii) the height of the tree.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q12.1
Let TR be the tree of height x m and y be the width AR of the river,
then ∠B = 30° and A = ∠60° , AB = 50 m.
Now in right ∆ATR,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q12.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q12.3

Question 13.
A 20 m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole, is 60° and the angle of elevation of the top of the pole as seen from the foot of the tower is 30°. Find
(i) the height of the tower.
(ii) the horizontal distance between the pole and the tower.
Solution:
Let PQ is the pole and TS is the tower. PQ = 20 m.
Let TS = h and QS = x Angles of elevation from Q to T A T is 60° and from S to P is 30°.
In the ∆PQS
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q13.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q13.2

Question 14.
A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole the angle of elevation of the top of the tower is 60° and the angle of depression of the bottom of the tower is 30°.
Find : (i) the height of the tower, if the height of the pole is 20m;
(ii) the height of the pole, if the height of the tower is 75 m.
Solution:
Let PQ is the pole and TW is the tower
Angle of elevation from T to P is 60° and angle of depression from P to W is 30°
∴ ∠PWQ = 30° = ∠RPW ( ∠ Altanate angles)
(i) In first case when height of pole OQ = 20m, Then in right ∆ PQW
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q14.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q14.2

Question 15.
From a point, 36 m above the surface of a lake, the angle of elevation of a bird is observed to be 30° and angle of depression of its image in the water of the lake is observed to be 60°. Find the actual height of the bird above the surface of the lake.
Solution:
Let AQ is the sea-level
P is a point 36 m above sea-level
∴ PQ = 36
Let B be the bird and R is its reflection in the water and angle of elevation of the bird B at P is 30° and angle of depression of the reflection of the bird at R is 60°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q15.2

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q16.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q16.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q16.3

Question 17.
As observed from the top of a 80 m tall lighthouse, the angles of depression of two ships, on the same side of the light house in horizontal line with its base , are 30° and 40° respectively . Find the distance between the two ships.Give your answer correct to the nearest meter. [2012]
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q17.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q17.2

Question 18.
In the figure given, from the top of a building AB = 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60° respectively. Find :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q18.1
(i) the horizontal distance between AB and CD.
(it) the height of the lamp post.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q18.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q18.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q18.4

Question 19.
An aeroplane at an altitude of 250 m observes the angle of depression of two boats on the opposite banks of a river to be 45° and 60° respectively. Find the width of the river. Write the answer correct to the nearest whole number. (2014)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q19.2

Question 20.
The horizontal distance between two towers is 120 m. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second tower is 30° and 24° respectively. Find the height of the two towers. Give your answer correct to 3 significant figures. (2015)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q20.1
Solution:
AB and CD are two towers which are 120 m apart
i.e. BD= 120m
Angles of elevation of the top and angle of depression of bottom of the first tower observed from the top of second tower is 30° and 24°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q20.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 22 Heights and Distances Ex 22C Q20.3

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E.

Other Exercises

Question 1.
Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio, 3 : 5. Find its co-ordinates of point P. Also, find the equation of the line through P and parallel to 3x + 5y = 7
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q1.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q1.2

Question 2.
The line segment joining the points A (3, -4) and B (-2, 1) is divided in the ratio 1 : 3 at point P in it Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x – 3y = 4.
Solution:
Point P, divides the line segment A (3, -4) and B(-2, 1) in the ratio of 1 : 3
Let co-ordinates of P be (x, y), then
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q2.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q2.2

Question 3.
A line 5x + 3y + 15 = 0 meets y -axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.
Solution:
P lies on y-axis and let the co-ordinates of P be (0, y)
P lies also on the line 5x + 3y + 15 = 0 it will satisfy it.
5 x 0 + 3y + 15 = 0
⇒ 3y = -15
⇒ y = -5
Co-ordinates of P are (0, -5)
Now, writing the line x – 3y + 4 = 0 is form of y = mx + c
-3y = -x – 4
⇒ 3y = x + 4
⇒ y = \(\frac { 1 }{ 3 }\) x + \(\frac { 4 }{ 3 }\)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q3.1

Question 4.
Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other. [2003]
Solution:
Writing, the line kx – 5y + 4 = 0 in form of y = mx + c
⇒ -5y = -kx – 4
⇒ 5y = kx + 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q4.1

Question 5.
A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid point of the segment AB. Find:
(i) the equation of the line.
(ii) the co-ordinates of A and B.
(iii) the co-ordinates of M. (2003)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q5.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q5.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q5.3
(ii) Let. co-ordinates of A be (x, 0) and of B be (0, y) which lie On the line.
Substituting, the co-ordinates in (i)
x + 0 = 3 ⇒x = 3
Co-ordinates of A are (3, 0)
Again 0 + y = 3 ⇒ y = 3
Co-ordinates of B are (0,3)
(iii) M is the mid-point of AB.
Co-ordinates of M wil be (\(\frac { 3 }{ 2 }\) , \(\frac { 3 }{ 2 }\))

Question 6.
(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.
Solution:
Co-ordinates of A and C of rhombus ABCD are (1, 5) and (-3, -1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q6.2
⇒ 3y – 6 = -2x – 2
⇒ 2x + 3y = 6 – 2
⇒ 2x + 3y = 4

Question 7.
Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
Solution:
Three vertices of a square ABCD are A (3, 2), B (6, -2) and C (2, -5)
Let, co-ordinates of fourth vertex D be (x, y)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q7.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q7.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q7.4

Question 8.
A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.
Solution:
Slope of line x = 3y + 2 or 3y = x – 2 ….(i)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q8.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q8.3

Question 9.
A straight line passes through the point (3, 2) and the portion of this line intercepted between the positive axes, is bisected at this point. Find the equation of the line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q9.1
Let, the line intersects x-axis at A and y-axis at B.
Let, co-ordinates of A (x, o) and of (o, y)
But (3, 2) is the mid-point of AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q9.2

Question 10.
Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x -2y = 1.
Solution:
7x + 6y = 71 ….(i)
5x – 8y = -23 ….(ii)
Multiply (i) by 4 and (ii) by 3,
28x + 24y = 284
15x – 24y = -69
On adding (i) and (ii), we get:
43x = 215
x = 5
Substituting, the value of x in (i)
7 x 5 + 6y = 71
35 + 6y = 71
⇒ 6y = 71 – 35 = 36
⇒ y = 6
Point of intersection of these lines is (5, 6)
Now slope of line 4x – 2y = 1
⇒ 4x – 1 = 2y
⇒ y = 2x – \(\frac { 1 }{ 2 }\) is 2
Slope of line through the point of intersection and perpendicular to 4x – 2y = 1 is \(\frac { -1 }{ 2 }\)
Equation of the line y – y1 = m (x – x1)
⇒ y – 6 = \(\frac { -1 }{ 2 }\) (x – 5)
⇒ 2y – 12 = -x + 5
⇒ x + 2y = 5 + 12 = 17
⇒ x + 2y = 17

Question 11.
Find the equation of the line which is perpendicular to the line \(\frac { x }{ a }\) – \(\frac { y }{ b }\) = 1 at the point where this line meets y-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q11.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q11.2

Question 12.
O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:
(i) the equation of median of ∆OAB through vertex O.
(ii) the equation of altitude of ∆OAB through vertex B.
Solution:
(i) Let, mid-point of AB be D.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q12.2

Question 13.
Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.
Does line 3x = y + 1 bisect the line segment joining the two given points ?
Solution:
Slope of the line joining the points (-2, 3) and (4, 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q13.1
Yes, these are perpendicular to each other
Let P be the mid-point of the line joining the points (-2, 3) and (4, 1)
Co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q13.2
This point (1, 2) satisfies the equaion 3x = y + 1 then, it will bisect the line joining the given point
now, substituting the value of x and y. in 3x = y + 1
⇒ 3 x 1 = 2 + 1
⇒ 3 = 3. which is true.
Yes, the line 3x = y + 1 is the bisector.

Question 14.
Given a straight line x cos 30° + y sin 30° = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).
Solution:
Equation of the given line is x cos 30° + y sin 30° = 2
y sin 36° = -x cos 30° + 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q14.1

Question 15.
Find the value of k such that the line (k – 2) x + (k + 3) y – 5 = 0 is :
(i) perpendicular to the line 2x – y + 7 = 0
(ii) parallel to it.
Solution:
Writing the given equation in the form of y = mx + c
(k – 2) x + (k + 3) y – 5 = 0 ….(i)
⇒ (k + 3) y = – (k – 2) x + 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q15.2

Question 16.
The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7) write down the equation of BC. Find :
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point P, where the perpendicular through A, as obtained in (i) meets BC.
Solution:
Vertices of ∆ABC are A (0, 5),B (-1, -2) and C (11, 7)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q16.1
(ii) Let, the line through A meets BC in P
P is point of intersection of these two lines.
3x – 4y = 5 ……… (i)
4x + 3y = 15 …….. (ii)
On solving (i), (ii) we get
x = 3, y = 1
Co-ordinates of Pare (3, 1)

Question 17.
From the given figure, find :
(i) the co-ordinates of A, B and C.
(ii) the equation of the line through A and parallel to BC. (2005)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q17.1
Solution:
(i) From the figure, we see that co-ordinates of A are (2, 3), of B are (-1, 2) of C and (3, 0)
(ii) Slope of line BC is (m)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q17.2
⇒ x + 2y – 6 – 2 = 0
⇒ x + 2y – 8 = 0
⇒ x + 2y = 8

Question 18.
P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R. (2004)
Solution:
Let (x, y) be the co-ordinates of M, the mid-point of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q18.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q18.2
⇒ x + 2y – 8 = 0
⇒ x + 2y = 8

Question 19.
A (8, -6), B (-4, 2) and C (0, -10) are the vertjces of a triangle ABC. If P is the mid-point of AR and Q is the mid-point of AC, use co ordinate geometry to show that PQ is parallel to BC. Give a special name to quadrilateral PBCQ.
Solution:
In ∆ABC, co-ordinates of A, B and C are (8, -6), (-4, 2) and C (0, -10) respectively.
P and Q are the mid-points of AB and AC respectively
Co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q19.2
Slopes of PQ and BC are same.
These are parallel to each other.
Quad. PBCQ is trapezium

Question 20.
A line AB meets the x-axis at point A and y-axis at point B. The point P(-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find :
(i) the co-ordinates of A and B.
(ii) equation of line through P, and perpendicular to AB.
Solution:
Line AB intersects x-axis at A and y-axis at B.
(i) Let co-ordinates of A be (x, 0) and of B be (0, y)
Point P (-4, -2) intersects AB in the ratio 1 : 2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q20.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q20.2

Question 21.
A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from die positive side of y-axis. Find the equation of the line. (1992)
Solution:
Let line intersects x-axis at P (-2, 0) and cuts off an intercept of 3 units at Q.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q21.1

Question 22.
Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units. (2002)
Solution:
x-intercept = 4
Co-ordinates of that point = (4, 0)
The co-ordinates of the given point (2, 3)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q22.1

Question 23.
The given figure (not drawn to scale) shows two straight lines AB and CD. If’ equation of the line AB is : y = x + 1 and equation of line CD is : y = √3 x – 1. Write down the inclination of lines AB and CD; also, find the angle 6 between AB and CD. (1989)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q23.1
Solution:
Equation of line AB is y = x + 1
and equation of line CD is y = √3 x – 1
Slope of AB = 1
tanθ = 1
⇒ θ = 45°
Inclination angles of AB = 45°
Slope of CD = tanθ = √3 = tan 60°
⇒ θ = 60°
Inclination angle of CD = 60°
In ΔPQR,
Ext. ∠RQX = ∠RPQ + ∠PRQ (Exterior angles is equal to sum of its interior opposite angles)
⇒ 60° = 45° + θ
⇒ θ = 60° – 45° = 15°
⇒ θ = 15°

Question 24.
Write down the equation of the line whose gradient is \(\frac { 3 }{ 2 }\) and which passes through P, where P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2 : 3. (1996, 2001)
Solution:
P divides the line segment AB in which A (-2, 6) and B (3, -4) in the ratio 2 : 3
Co-ordinates of P will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q24.1

Question 25.
The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.
Solution:
Let points are A (6, 4) and B (7, -5)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q25.1

Question 26.
Points A and B have coordinates (7, -3) and (1, 9) respectively. Find
(i) the slope of AB.
(ii) the equation of the perpendicular bisector of the line segment AB.
(iii) the value of ‘p’ of (-2, p) lies on it.
Solution:
Coordinates of A are (7, -3), of B = ( 1, 9)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q26.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q26.2

Question 27.
A and B are two points on the x-axis and y-axis respectively. P (2, -3) is the mid point of AB. Find the
(i) Coordinates of A and B.
(ii) Slope of line AB.
(iii) equation of line AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q27.1
Solution:
As P (2, -3) is mid-point of AB.
Let coordinates of B be (0, y) and coordinates of A be (x, 0)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q27.2

Question 28.
The equation of a line is 3x + 4y – 7 = 0. Find:
(i) the slope of the line.
(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.
Solution:
Given line 3x + 4y -7 = 0
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q28.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q28.2
⇒ 3 (y – 4) = 4 (x – 2)
⇒ 3y – 12 = 4x – 8
⇒ 4x – 3y – 8 + 12 = 0
⇒ 4x – 3y + 4 = 0

Question 29.
ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4). Find :
(i) co-ordinates of A
(ii) equation of diagonal BD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q29.1
(i) In || gm ABCD, A (x, y), B (5, 8), C (4, 7) and D (2, -4)
The diagonals of ||gm bisect each other
O is said point of AC and BD
Now if O is mid point of BD then its co-ordinates will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q29.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q29.3
⇒ y + 4 = 4x – 8
⇒ 4x – y -8 – 4 = 0
⇒ 4x – y – 12 = 0 or 4x – y = 12

Question 30.
Given equation of line L1 is y = 4.
(i) Write the slope of line L2 if L2 is the bisector of angle O.
(ii) Write the co-ordinates of point P.
(iii) Find the equation of L2.
Solution:
(i) Equation of line L1 is y = 4
L2 is the bisector of ∠O
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q30.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q30.2

Question 31.
Find:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q31.1
(i) equation of AB
(ii) equation of CD
Solution:
Co-ordinates of A and B are (-5, 4) and (3, 3) respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q31.2

Question 32.
Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1
Solution:
x-intercept of the line = -3
and is perpendicular to the line
3x + 5y = 1
5y = 1 – 3x
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q32.1

Question 33.
A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid-point of the line segment AB. Find :
(i) the equation of the line.
(ii) the co-ordinates of points A and B.
(iii) the co-ordinates of point M.
Solution:
A line passing through the two points P (-1, 4) and Q (5, -2) intersects x-axis at point A and y- axis at point B.
M is mid-point of AB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q33.1
(i) Equation of line AB will be
y – y1 = m (x – x1)
⇒ y – 4 = -1 (x + 1)
⇒ y – 4 = – x – 1
⇒ y + x = -1 + 4
⇒ x + y = 3
(ii) The line intersect x-axis at A and OA = 3 units
Co-ordinates of A are (3, 0) and the line intersects y-axis at B and OB = 3 units
Co-ordinates of B are (0, 3)
(iii) M is mid-point of AB
Co-ordinates of M are (\(\frac { 3 }{ 2 }\) , \(\frac { 3 }{ 2 }\))

Question 34.
In the given figure, line AB meets y-axis at point A. Line through C (2, 10) and D intersects line AB at right angle at point P. Find:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q34.1
(i) equation of line AB.
(ii) equation of line CD.
(iii) co-ordinates of points E and D.
Solution:
In the given figure, AB meets y-axis at point A.
Line through C (2, 10) and D intersects line AB at P at right angle.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q34.2
Equation of CD
y – 10 = 3 (x – 2)
⇒ y – 10 = 3x – 6
⇒ 3x – y + 10 – 6 = 0
⇒ 3x – y + 4 = 0
(iii) Co-ordinates of D which is on x-axis
3x – y + 4 = 0
3x – 0 + 4 = 0
⇒ 3x + 4 = 0
⇒ 3x = -4
⇒ x = \(\frac { -4 }{ 3 }\)
Co-ordinates of D are (\(\frac { -4 }{ 3 }\) , 0)
E is also on x-axis
x + 3y = 18
Substituting, y = 0, then
x + 0= 18
⇒ x = 18
Co-ordinates of E are (18, 6)

Question 35.
A line through point P (4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.
Solution:
A line through P (4, 3) meets x-axis at A and the y-axis at B. If BP is double of PA.
Draw BC || x-axis
and PC || y-axis
In ∆PAD and ∆PBC
∠P = ∠P (common)
∠D = ∠C (each 90°)
∆PAD ~ ∆PBC
PB = 2PA ⇒ PA = \(\frac { 1 }{ 2 }\) PB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q35.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q35.2
2y – 6 = 3x – 12
⇒ 3x – 2y – 12 + 6 = 0
⇒ 3x – 2y – 6 = 0
⇒ 3x – 2y = 6

Question 36.
Find the equation of line through the intersection of lines 2x – y = 1 and 3x + 2y = -9 and making an angle of 30° with positive direction of x-axis.
Solution:
Equation of given two intersecting lines are 2x – y = 1 and 3x + 2y = -9 Which make an angle of 30°
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q36.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q36.2

Question 37.
Find the equation of the line through the points A (-1, 3) and B (0, 2). Hence, show that the points A, B and C (1, 1) are collinear.
Solution:
The given points are A (-1, 3) and B (0, 2) and co-ordinates of a point C are (1, 1)
Now slope of the line joining A and B
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q37.1
Equation of the line y – y1 = m(x – x1)
⇒ y – 2 = -1 (x – 0)
⇒ y – 2 = -x
⇒ x + y – 2 = 0
Point C (1, 1) will be on AB if it satisfy
1 + 1 – 2 = 0
⇒ 0 = 0
Point C lies on AB
Hence points A, C and B are collinear.

Question 38.
Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2), find:
(i) the co-ordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD. (2015)
Solution:
Three vertices of a ||gm ABCD taken an order are A (3, 6), B (5, 10) and C (3, 2)
Join diagonals AC and BD which bisect each other at O.
O is mid-point of AC as well as of BD
Now co-ordinates of O will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q38.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q38.2

Question 39.
In the figure, given, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q39.1
(i) Write the co-ordinates of A.
(ii) Find the length of AB and AC.
(iii) Find the ratio in which Q divides AC.
(iv) Find the equation of the line AC. (2015)
Solution:
In the given figure,
ABC is a triangle and BC || y-axis
AB and AC intersect the y-axis at P and Q respectively.
(i) Co-ordinates of A are (4,0).
(ii) Length of AB
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q39.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q39.3

Question 40.
The slope of a line joining P (6, k) and Q (1 – 3k, 3) is \(\frac { 1 }{ 2 }\). Find :
(i) k
(ii) mid-point of PQ, using the value of ‘A’ found in (i). (2016)
Solution:
(i) Slope of the line joining P(6, k) and Q (1 –
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q40.1

Question 41.
A line AB meets X-axis at A and Y-axis at B. P (4, -1) divides AB in the ratio 1 : 2.
(i) Find the coordinates of A and B.
(ii) Find the equation of the line through P and perpendicular to AB.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q41.1
Solution:
(i) Since, A lies on the x-axis,
let the coordinates of A be (x, 0).
Since B lies on the y-axis,
let the coordinates of B be (0, y).
Let m = 1 and n = 2.
Using section formula,
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14E Q41.2

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D.

Other Exercises

Question 1.
Find the slope and y-intercept of the line :
(i) y = 4
(ii) ax – by = 0
(iii) 3x – 4y = 5
Solution:
(i) y = 4 ⇒ y = 0x + 4
Here slope = 0 and y-intercept = 4
(ii) ax – by = 0
⇒ by = ax
⇒ y = \(\frac { a }{ b }\) x + 0
Here, slope = \(\frac { a }{ b }\) and y-intercept = 0
(iii) 3x – 4y = 5
⇒ – 4y = 5 – 3x
⇒ 4y = 3x – 5
⇒ y = \(\frac { 3 }{ 4 }\) x + \(\frac { -5 }{ 4 }\)
Here, slope = \(\frac { 3 }{ 4 }\) and y- intercept = \(\frac { -5 }{ 4 }\)

Question 2.
The equation of a line is x – y = 4. Find its – slope and y-intercept. Also, find its inclination.
Solution:
x – y = 4
writing the equation in form of y = mx + c
x = 4 + y
⇒ y = x – 4
Slope = 1 and y-intercept = – 4
Slope = 1
⇒ tanθ = 1
⇒ θ = 45°

Question 3.
(i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0 ?
(ii) Is the line x – 3y = 4 perpendicular to the line 3x – y = 7 ?
(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1 ?
(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.
Solution:
(i) 3x + 4y + 7 = 0
Writing the equation in form of y = mx + c
4y = -3x – 7
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q3.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q3.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q3.4

Question 4.
Find the slope of the line which is parallel to:
(i) x + 2y + 3 = 0
(ii) \(\frac { x }{ 2 }\) – \(\frac { y }{ 3 }\) – 1 = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q4.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q4.2

Question 5.
Find the slope of the line which is perpendicular to:
(i) x – \(\frac { y }{ 2 }\) + 3 = 0
(ii) \(\frac { x }{ 3 }\) – 2y = 4
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q5.2

Question 6.
(i) Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.
(ii) Lines mx + 3y + 7 = 0 and 5x- ny – 3 = 0 are perpendicular to each other. Find the relation connecting m and n.
Solution:
(i) Writing the given equations in the form of y = mx + c, we get:
-by = -2x -5
by = 2x + 5
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q6.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q6.2

Question 7.
Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.
Solution:
Writing the given equations 2x – y + 5 = 0 and px + 3y = 4 in form of y = mx + c
2x – y + 5 = 0
-y = -2x -5
y = 2x + 5
Here, slope of the line = 2
Again, px + 3y = 4
3y = – px + 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q7.1

Question 8.
The equation of a line AB is 2x – 2y + 3 = 0.
(i) Find the slope of the line AB.
(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.
Solution:
The line AB is given.2* – 2y + 3 = 0
Writing it in the form of y = mx + c
-2y = -2x – 3
⇒ 2y = 2x + 3
⇒ y = x + \(\frac { 3 }{ 2 }\)
Here, slope of the line = 1
Angle of inclination = tanθ
tanθ = 1
θ = 45°

Question 9.
The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.
Solution:
Writing the given lines 4x + 3y = 9 and px – 6y + 3 = 0 in the form of y = mx + c
4x + 3y = 9
⇒ 3y = – 4x + 9
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q9.1

Question 10.
If the lines y = 3x + 7 and 2y +px = 3 are perpendicular to each other, find the value of p. (2006)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q10.1

Question 11.
The line through A (-2, 3) and B (4, 6) is perpendicular to the line 2x – 4y = 5. Find the value of b.
Solution:
Gradient (mx) of the line passing through the points A (-2, 3) and B (4, b)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q11.1

Question 12.
Find the equation of the line passing through (-5, 7) and parallel to
(i) x-axis
(ii) y-axis.
Solution:
(i) Slope of the line parallel to x-axis = 0
Equation of line passing through (-5, 7) whose slope is 0.
y – 7 = 0 [x – (-5)]
⇒ y – 7 = 0
⇒ y = 7
(ii) Slope of the line parallel to y-axis = 0
y – y1 = m (x – x1)
⇒ 0 = x – x1
⇒ x + 5 = 0

Question 13.
(i) Find the equation of the line passing through (5, -3) and parallel to x – 3y = 4.
(ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1). (2007)
Solution:
(i) Writing the equation x – 3y = 4 in form of y = mx + c
-3y = -x + 4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q13.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q13.2
⇒ 2y – 2 = -3x
⇒ 3x + 2y – 2 = 0
Which is the required equation.

Question 14.
Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.
Solution:
Writing the equation 4x + 5y = 6 in form of y = mx + c
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q14.1

Question 15.
Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).
Solution:
The perpendicular of the line segment bisects it.
Co-ordinates of mid-point of the line segment which is obtained by joining the points (6, -3) and (0, 3)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q15.1
Slope of the line perpendicular to it = 1 (Product of slopes = -1)
Equation of the perpendicular bisector is y – y1 = m (x – x1)
y – 0 = 1 (x – 3)
y = x – 3

Question 16.
In the following diagram, write down:
(i) the co-ordinates of the points A, Band C.
(ii) the equation of the line through A and parallel to BC.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q16.1
Solution:
(i) From the figure, the see that co-ordinates of A are (2, 3), B are (-1, 2) and C are (3, 0)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q16.2

Question 17.
B (-5, 6) and D (1, 4) are the vertices of rhombus ABCD. Find the equation of diagonal BD and of diagonal AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q17.1

Question 18.
A = (7, -2) and C = (-1, – 6) are the vertices of a square ABCD. Find the equation of the diagonals AC and BD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q18.1

Question 19.
A (1, -5), B (2, 2) and C (-2, 4) are the vertices of the ∆ABC, find the equation of:
(i) the median of the triangle through A.
(ii) the altitude of the triangle through B.
(iii) the line through C and parallel to AB.
Solution:
(i) Let D be the mid-point of BC
co-ordinates mid-point of
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q19.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q19.3

Question 20.
(i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.
(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin. [1995]
Solution:
(i) Write the equation 2y = 3x + 5 in the form of y = mx + c. We get:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q20.1
y – 2 = \(\frac { -2 }{ 3 }\) (x – 3)
⇒ 3y – 6 = -2x + 6
⇒ 2x + 3y = 6 + 6
⇒ 2x + 3y = 12 …… (i)
(ii) AB meets the x-axis at A
ordinate (y) of A = 0 i.e. y = 0
Substituting, the value of y in (i)
2x + 3 x 0 = 12
⇒ 2x = 12
⇒ x = 6
Co-ordinates of A are (6, 0)
Again. AB meets y-axis at B
Abscissa of B = 0 i.e. x = 0
Substituting the value of x in (i)
2 x 0 + 3y = 12
⇒ y = 4
Co-ordinates of B are (0, 4)
Area of ∆OAB = \(\frac { 1 }{ 2 }\) x Base x altitude
= \(\frac { 1 }{ 2 }\) x 4 x 6 = 12 square units

Question 21.
The line 4x – 3y + 12 = 0 meets the x-axis at A. Write the co-ordinates of A.
Determine the equation of the line through A and perpendicular to 4x – 3y + 12 = 0.
Solution:
The line 4x – 3y + 12 = 0 meets x-axis at A.
Ordinates of A = 0. i.e. y = 0
Substituting, the value of y in the equation
4x – 3 x 0 + 12 = 0
⇒ 4x + 12 = 0
⇒ 4x = -12
⇒ x = -3
Co-ordinates of A are (-3, 0)
Writing the equation 4x – 3y + 12 = 0 in form of y = mx + c
⇒ -3y = -4x – 12
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q21.1

Question 22.
The point P is the foot of perpendicular from A (-5, 7) to the line is 2x – 3y + 18 = 0. Determine:
(i) the equation of the line AP.
(ii) the co-ordinates of P.
Solution:
Write the equation in form of y = mx + c
2x – 3y + 18 = 0
⇒ -3y = -2x – 18
⇒ y = \(\frac { 2 }{ 3 }\) x + 6 (Dividing by 3)
Slope of the line = \(\frac { 2 }{ 3 }\)
and slope of the line perpendicular to it = \(\frac { -3 }{ 2 }\) (Product of slopes = -1)
(i) Equation of line AP perpendicular to the given line and drawn through A (-5, 7)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q22.1
y – y1 = m (x – x1)
⇒ y – 7 = \(\frac { -3 }{ 2 }\) (x + 5)
⇒ 2y – 14 = -3x – 15
⇒ 3x + 2y – 14 + 15 = 0
⇒ 3x + 2y + 1 = 0
(ii) P is the point of intersection of these lines
we will solve their equations
2x – 3y + 18 = 0 ….(i)
3x + 2y + 1 = 0 ….(ii)
Multiplying (i) by 2 and (ii) by 3, we get
4x – 6y + 36 = 0
9x + 6y + 3 = 0
Adding, we get:
13x + 39 = 0
⇒ 13x = -39
⇒ x = -3
Now, substituting the value of x in (i)
2(-3) – 3y + 18 = 0
⇒ -6 – 3y + 18 = 0
⇒ -3y + 18 – 6 = 0
⇒ -3y + 12 = 0
⇒ -3y = -12
⇒ 3y = 12
⇒ y = 4
Co-ordinates of P are (-3, 4)

Question 23.
The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC. If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the co-ordinates of P and Q.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q23.1
AB meets y- axis at P
abscissa of P = 0 i. e. x = 0
Substituting the value of y in (i)
0 + y = 4
⇒ y = 4
Co-ordinates of P are (0, 4)
BC meets x-axis at Q
ordinate of Q = 0 i.e. y = 0
Substituting, the value of y in (ii),
2x + 0 = 6
⇒ 2x = 6
⇒ x = 3
Co-ordinates of Q are (3, 0)

Question 24.
Match the equations A, B, C, and D with the lines L1, L2, L3 and L4, whose graphs are roughly drawn in the given diagram.
A = y = 2x;
B = y – 2x + 2 = 0;
C = 3x + 2y = 6;
D = y = 2 [1996]
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q24.1
Solution:
A → L3,
B → L4,
C → L2,
D → L1

Question 25.
Find the value of ‘a’ for which the following points A (a, 3), B (2, 1) and C (5, a) are collinear. Hence find the equation of the line. (2014)
Solution:
A (a, 3), B (2,1) and C (5, a) are collinear.
Slope of AB = Slope of BC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D Q25.1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C.

Other Exercises

Question 1.
Find the equation of line whose :
(i) y-intercept = 2 and slope = 3,
(ii) y-intercept = -1 and slope = \(\frac { -3 }{ 4 }\)
Solution:
(i) The point whose y-intercept = 2 will be (0, 2) and slope (m) = 3.
Equation of line will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q1.1

Question 2.
Find the equation of a line whose :
(i) y-intercept = -1 and inclination = 45°
(ii) y-intercept = 3 and inclination = 30°
Solution:
(i) The point whose y-intercept is -1, will be (0, -1) and inclination = 45°
Slope (m) = tan 45° = 1
Equation will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q2.1

Question 3.
Find the equation of the line whose slope is \(\frac { -4 }{ 3 }\) and which passes through (-3, 4).
Solution:
Slope of the line (m) = \(\frac { -4 }{ 3 }\)
The point from which the line passes (-3, 4)
Equation of line will be y – y1 = m (x – x1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q3.1

Question 4.
Find the equation of a line which passes through (5, 4) and makes an angle of 60° with the positive direction of the x-axis.
Solution:
The line passes through the point (5, 4) and angle of inclination = 60°
slope (m) = tan 60° = √3
Equation of line
y – y1 = m (x – x1)
⇒ y – 4 = √3 (x – 5)
⇒ y – 4 = √3 x – 5√3
⇒ y = √3 x + 4 – 5√3

Question 5.
Find the equation of the line passing through:
(i) (0, 1) and (1, 2)
(ii) (-1, -4) and (3, 0)
(iii) (4, -2) and (5, 2)
Solution:
Two given points are (0, 1) and (1, 2)
Slope of the line passing through these two
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q5.2

Question 6.
The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find :
(i) The gradient of PQ
(ii) The equation of PQ,
(iii) The co-ordinates of the point where PQ intersects the x-axis.
Solution:
Two points P (2,-6) and Q (-3, 5) are given.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q6.1
⇒ 5y – 30 = x – 2
⇒ 5y = x – 2 + 30
⇒ 5y = x + 28 ….(i)
(iii) Co-ordinates of the point where PQ intersects x-axis will be = 0
substituting, the value of y in (i)
5 x 0 = x + 28 ⇒ x + 28 = 0 ⇒ x = -28
Co-ordinates of point are (-28, 0)

Question 7.
The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find :
(i) the equation of AB
(ii) the co-ordinates of the point where the line AB intersects they- axis.
Solution:
Slope of the line passing through two points A (-3, 4) and B (2, -1) will be :
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q7.1
Its abscissa = 0
substituting, the value of x = 0 in (i)
0 + y = 1
y = 1
Co-ordinates of point = (0, 1)

Question 8.
The figure given below shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equations of AB and CD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q8.1
Solution:
Two lines AB and CD intersect each other at P (3, 4)
AB inclined at angle of 45° and CD at angle of 60° with the x-axis.
Slope of AB = tan 45° = 1
and slope of CD = tan 60° = √3
Now, equation of line AB will be
y – y1 = m (x – x1)
⇒ y – 4 = 1 (x – 3)
⇒ y – 4 = x – 3
⇒ y = x – 3 + 4
⇒ y = x + 1
(ii) Equation of CD will be :
y – y1 = m (x – x1)
⇒ y – 4 = √3 (x – 3)
⇒ y – 4 = √3 x – 3√3
⇒ y = √3 x – 3√3 + 4
⇒ y = √3 x + 4 – 3√3

Question 9.
In ΔABC, A (3, 5), B (7, 8) and C (1, -10). Find the equation of the median through A.
Solution:
AD is median
D is mid point of BC
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q9.2
Equation of AD
y – y1 = m (x – x1)
⇒ y + 1 = -6 (x – 4)
⇒ y + 1 = -6x + 24
⇒ y + 6x = -1 + 24
⇒ 6x + y = 23

Question 10.
The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, ∠A = 60° and vertex, C = (7, 5). Find the equations of BC and CD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q10.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q10.2
ABCD is a ||gm in which AB = CD || x-axis
∠A = 60° and C (7, 5)
(i) CD || AB || x-axis ,
Equation of CD will be
y – y1 = m (x – x1)
⇒ y – 5 = 0 (x – 7)
⇒ y – 5 = 0
⇒ y = 5
BC || AD
Slope of BC = tan 60° = √3
Equation of BC will be
y – y1 = m (x – x1)
⇒ y – 6 = √3 (x – 7)
⇒ y – 6 = √3 x – 7√3
⇒ y = √3 x + 6 – 7√3

Question 11.
Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x – y = 4.
Solution:
Point of intersection of two lines
x + 2y = 7 ….(i)
x – y = 4 ….(ii)
Subtracting, we get
3y = 3
y =1
Substituting, the value of y in (ii)
x – 1 = 4
⇒ x = 4 + 1 = 5
Point of intersection is (5, 1)
Slope of the line passing through origin (0, 0) and (5, 1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q11.1
Equation of line will be
y – y1 = m (x – x1)
⇒ y – 5 = \(\frac { 1 }{ 2 }\) (x – 1)
⇒ 5y – 25 = x – 1
⇒ 5y = x – 1 + 25 = x + 24
⇒ 5y = x + 24

Question 12.
In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A. Also, find the equation of the line through vertex B and parallel to AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q12.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q12.2

Question 13.
A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q13.1
Slope of line through A, perpendicular to BC = -(-1) = 1
Now, the equation of line through A (0, 3) is
y – y1 = m (x – x1)
y – 3 = 1 (x – 0)
⇒ y – 3 = x
⇒ y = x + 3

Question 14.
Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).
Solution:
Slope of the line joining the points (1, 4) and (2, 3)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q14.1
Slope of line perpendicular to the above line = 1
Equation of line passing through (-1, 2)
y – y1 = m (x – x1)
⇒ y – 2 = 1 [x -(-1)]
⇒ y – 2 = x + 1
⇒ y = x + 1 + 2
⇒ y = x + 3

Question 15.
Find the equation of the line, whose :
(i) x-intercept = 5 and y-intercept = 3
(ii) x-intercept = -4 and y-intercept = 6
(iii) x-intercept = -8 and y-intercept = -4
(iv) x-intercept = 3 and y-intercept = -6
Solution:
(i) When x-intercept = 5, then point will be (5, 0)
and when y-intercept = 3, then point will be (0, 3)
Slope of the line passing through these points
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q15.3

Question 16.
Find the equation of the line whose slope is \(\frac { -5 }{ 6 }\) and x-intercept is 6.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q16.1

Question 17.
Find the equation of the line with x-intercept 5 and a point on it (-3, 2).
Solution:
x-intercept of the line = 5
Point = (5, 0)
Slope of the line passing through the point (-3, 2)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q17.1

Question 18.
Find the equation of the line through (1, 3) and making an intercept of 5 on the y- axis.
Solution:
The line makes y-intercept = 5
Point = (0, 5)
Slope of the line passing through the point (1, 3) and (0, 5)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q18.1
Equation of the line
y – y1 = m (x – x1)
⇒ y – 3 = -2 (x – 1)
⇒ y – 3 = -2x + 2
⇒ 2x + y = 2 + 3
⇒ 2x + y = 5

Question 19.
Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axes.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q19.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q19.2
Solution:
(i) Slope of line AB = tan 45° = 1
Equation passing through the point (-2, 0)
y – y1 = m (x – x1)
⇒ y – 0 = 1 (x + 2)
⇒ y = x + 2
⇒ x – y + 2 = 0
(ii) Slope of line CD = tan (-45°) = -1
Equation passing through the point (-2, 0)
y – y1 = m (x – x1)
⇒ y – 0 = -1 (x + 2)
⇒ y = -x – 2
⇒ y + x + 2 = 0
⇒ x + y + 2 = 0

Question 20.
The line through P (5, 3) intersects y axis at Q.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q20.1
(i) Write the slope of the line.
(ii) Write the equation of the line.
(iii) Find the co-ordinates of Q.
Solution:
(i) Here θ = 45°
So, slope of the line = tanθ = tan 45° = 1
(ii) Equation of the line through P and Q is
y – 3 = 1 (x – 5)
⇒ y – x + 2 = 0
(iii) Let the coordinates of Q be (0, y)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q20.2

Question 21.
Write down the equation of the line whose gradient is \(\frac { -2 }{ 5 }\) and which passes through point P, where P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3 : 1.
Solution:
Slope of the line m = \(\frac { -2 }{ 5 }\)
P divides the line AB, whose co-ordinates are (4, -8) and (12, 0) in the ratio of 3 : 1
Co-ordinates of P be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q21.1

Question 22.
A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC. Find :
(i) the co-ordinates of the centroid of ΔABC.
(ii) the equation of a line, through the centroid and parallel to AB. [2002]
Solution:
(i) Co-ordinates of vertices of ΔABC are A (1, 4), B (3, 2), C (7, 5)
and let G be the centroid of ΔABC.
Co-ordinates of G are
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q22.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q22.2

Question 23.
A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point F in AC; such that AP : CP = 2 : 3.
Solution:
P divides AC in the ratio of 2 : 3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q23.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14C Q23.2

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If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B.

Other Exercises

Question 1.
Find the slope of the line whose inclination is :
(i) 0°
(ii) 30°
(iii) 72° 30′
(iv) 46°
Solution:
(i) Slope of line whose inclination is 0° = tanθ = tan 0° = 0
(ii) Slope of line whose inclination is 30° = tan 30° = \(\frac { 1 }{ \surd 3 }\)
(iii) Slope of line whose inclination is 72° 30′ = tan 72°30’ = 3. 1716 (Using tables)
(iv) Slope of line whose inclination is 46° = tan 46° = 1.0355 (Using tables)

Question 2.
Find the inclination of the line whose slope is:
(i) 0
(ii) √3
(iii) 0.7646
(iv) 1.0875
Solution:
Slope of a line = tanθ. Where θ is the inclination
(i) When slope is θ. then tanθ = 0 ⇒ θ = 0°
(ii) When slope is θ, then tanθ = √3 ⇒ θ = 60°
(iii) When slope is 0.7646, then tanθ = 0.7646 ⇒ θ = 37°24′ (Using tables)
(iv) When slope is 1.0875, then tanθ = 1.0875 ⇒ θ = 47°24′ (Using tables)

Question 3.
Find the slope of the line passing through the following pairs of points :
(i) (-2, -3) and (1, 2)
(ii) (-4, 0) and origin
(iii) (a, -b ) and (b, -a)
Solution:
We know that, slope of a line which passes
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q3.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q3.2

Question 4.
Find the slope of the line parallel to AB if:
(i) A = (-2, 4) and B = (0, 6)
(ii) A = (0, -3) and B = (-2, 5)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q4.1

Question 5.
Find the slope of the line perpendicular to AB if:
(i) A = (0, -5) and B = (-2, 4)
(ii) A = (3, -2) and B = (-1, 2)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q5.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q5.2

Question 6.
The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.
Solution:
Slope of the line passing through two points (0, 2) and (-3, -1)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q6.1

Question 7.
The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.
Solution:
Slope of the line passing through the points
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q7.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q7.2

Question 8.
Without using the distance formula, show that the point A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q8.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q8.2
AB and CA are perpendicular to each other
Hence, ΔABC is a right-angled triangle.

Question 9.
Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q9.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q9.2
Slopes of AB and DC are equal
AB || DC Similarly slope of BC and slope of DA are equal.
BC || DA
Hence ABCD is a parallelogram.

Question 10.
(-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the – quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q10.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q10.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q10.3
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q10.4
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q10.5
QR || PS.
Hence PQRS is a parallelogram.

Question 11.
Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are collinear.
Solution:
The given points are P (a, b + c), Q (b, c + a) and R (c, a + b)
We know that, these points P, Q, R are collinear if Slope of PQ = Slope of QR
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q11.1
Slope of PQ = Slope of QR
P, Q and R are collinear.

Question 12.
Find x, if the slope of the line joining (x, 2) and (8, -11) is \(\frac { -3 }{ 4 }\).
Solution:
Slope of line joining (x, 2) and (8, -11) is
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q12.1

Question 13.
The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slopes of all its sides.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q13.1
Solution:
ΔABC is an equilateral
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q13.2
Each angle is equal to 60°
Side AB is parallel to x-axis
Slope of AB = slope of x-axis = 0.
Slope of AC = tan A = tan 60° = √3
Slope of CB = tan B = tan 120° = tan (180°- 60°) = – tan 60° = -√3
Slopes of AB, BC and CA are 0, -√3, √3

Question 14.
The side AB of a square ABCD is parallel to the x-axis. Find the slopes of all its sides. Also, find :
(i) the slope of the diagonal AC
(ii) the slope of the diagonal BD.
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q14.1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q14.2
ABCD is a square in which AB || DC || x-axis.
AD || BC || y-axis
Slope of AB and DC = 0
and slope of AD and BC = not defined (tan90° is not defined)
AC and BD are the diagonals of square ABCD.
Now slope of AC = tan 45° = 1
and slope of BD = tan 135° = tan (180° – 45°) = – tan 45° = -1

Question 15.
A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find :
(i) the slope of the altitude of AB
(ii) the slope of the median AD and
(iii) the slope of the line parallel to AC.
Solution:
Vertices of ΔABC are A (5, 4), B (-3, -2), and C (1, -8)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q15.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q15.2
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q15.3

Question 16.
The slope of the side BC of a rectangle ABCD is \(\frac { 2 }{ 3 }\). Find
(i) The slope of the side AB,
(ii) the slope of the side AD.
Solution:
ABCD is a rectangle in which
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q16.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q16.2

Question 17.
Find the slope and the inclination of the line AB if
(i) A = (-3, -2) and B = (1, 2)
(ii) A = (0, -√3) and B = (3, 0)
(iii) A = (-1, 2√3) and B = (-2, √3)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q17.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q17.2

Question 18.
The points (-3, 2), (2, -1) and (a, 4) are collinear, Find ‘a’.
Solution:
Points are collinear.
Slope of (-3, 2) and (2, -1) = Slope of (2, -1) and (a, 4)
Now, Slope of (-3, 2) and (2, -1) will be
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q18.1

Question 19.
The points (k, 3), (2, -4) and (-k + 1, -2) are collinear. Find k.
Solution:
Points (k, 3), (2, -4) and (-k + 1, -2) are collinear
Slope of (k, 3) and (2, -4) = slope of (2, -4) and (-k + 1, -2)
Now, slope of (k, 3) and (2, -4)
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q19.1

Question 20.
Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C.
Which segment appears to have the steeper slope, AB or AC ?
Justify your conclusion by calculating the slopes of AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q20.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q20.2

Question 21.
Find the value(s) of k so that PQ will be parallel to RS. Given :
(i) P (2, 4), Q (3, 6), R (8, 1) and S (10, k)
(ii) P (3, -1), Q (7, 11), R (-1, -1) and S (1, k)
(iii) P (5, -1), Q (6, 11), R (6, -4k) and S (7, k²)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q21.1
Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B Q21.2

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line Ex 14B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.