Selina ICSE Solutions for Class 10 Maths

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

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 14A.

Other Exercises

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

Question 1.
Find, which of the following points lie on the line x – 2y + 5 = 0 :
(i) (1, 3)
(ii) (0, 5)
(iii) (-5, 0)
(iv) (5, 5)
(v) (2, -1.5)
(vi) (-2, -1.5)
Solution:
Equation of given line x – 2y + 5 = 0
(i) Substituting x = 1, y = 3, in the given equation.
1 – 2 x 3 + 5 = 0 ⇒ 1 – 6 + 5 = 0 ⇒ 0 = 0, which is true.
(1, 3) satisfies the equation.
(ii) Substituting x = 0 , y = 5 in the given equation
0 – 2 x 5 + 5 = 0 ⇒ 0 – 10 + 5 = 0 ⇒ -5 = 0, which is not true.
( 0, 5) does not satisfy the equation.
(iii) Substituting x = – 5, y = 0 in the given equation
-5 – 2 x 0 + 5 = 0 ⇒ -5 – 0 + 5 = 0 ⇒ 0 = 0 which is true.
(-5, 0) satisfies the equation.
(iv) Substituting x = 5, y = 5 in the given equation.
– 5 – 2 x 5 + 5 = 0 ⇒ -5 – 10 + 5 = 0 ⇒ 0 = 0 which is true.
(5, 5) satisfies the equation.
(v) Substituting x = 2, y = -1.5 in the given equation.
2 – 2 x (- 1.5) + 5 = 0 ⇒ 2 + 3 + 5 = 0 ⇒ 10 = 0. which is not true.
(2, -1.5) does not satisfy the equation.
(vi) Substituting x = -2, y = -1.5 in the given equation
– 2 – 2 x (-1.5) + 5 = 0 ⇒ – 2 + 3 + 5 = 0 ⇒ 6 = 0, which is not true.
(-2, -1.5) does not satisfies the equation.

Question 2.
State, true or false :
(i) the line \(\frac { x }{ 2 }\) + \(\frac { y }{ 3 }\) = 0 passes through the point (2, 3).
(ii) the line \(\frac { x }{ 2 }\) + \(\frac { y }{ 3 }\) = 0 passes through the point (4, -6).
(iii) the point (8, 7) lies on the line y – 7 = 0
(iv) the point (-3, 0) lies on the line x + 3 = 0
(v) if the point (2, a) lies on the line 2x – y = 3, then a = 5.
Solution:
(i) Equation of the line is \(\frac { x }{ 2 }\) + \(\frac { y }{ 3 }\) = 0
and co-ordinates of point are (2, 3)
If the point is on the line, then it will satisfy the equation.

(2, 3) is not on the line
(ii) Equation of the line is \(\frac { x }{ 2 }\) + \(\frac { y }{ 3 }\) = 0
and co-ordinates of point are (4, -6)
If the point is on the line, then it will satisfy the equation

Hence, point (4, -6) is on the line.
(iii) Equation of line is y – 7 = 0 and the co-ordinates of point are (8, 7)
If the point is on the line, then it will satisfy the equation
L.H.S. = y – 7 = 7 – 7 = 0 = R.H.S.
Hence, point (8, 7) is on the line.
(iv) Equation of the line is x + 3 = 0 and co-ordinates of point are (-3, 0)
If the point is on the line, then it will satisfy the equation.
L.H.S. = x + 3 = -3 + 3 = 0 = R.H.S.
Hence, the point (-3, 0) is on the line.
(v) Equation of the line is 2x – y = 3
and co-ordinates of the point are (2, a)
If the point is on the line, then it will satisfy the equation.
L.H.S. = 2x – y = 2 x 2 – a = 4 – a
R.H.S. = 3
4 – a = 3 ⇒ 4 + 3 = a ⇒ a = 7
But a = 5 given, therefore it is not on the line.
(i) False (ii) True (iii) True (iv) True (v) False.

Question 3.
The line given by the equation 2x – \(\frac { y }{ 3 }\) = 7 passes through the point (k, 6); calculate the value of k.
Solution:

Question 4.
For what value of k will the point (3, -k) lie on the line 9x + 4y = 3 ?
Solution:
Point (3, -k) satisfies the equation 9x + 4y = 3
Substituting x = 3 , y = -k, we get :
9 x 3 + 4 (- k), = 3
⇒ 27 – 4k = 3
⇒ – 4k = 3 – 27
⇒ – 4k = – 24
⇒ k = 6

Question 5.
The line \(\frac { 3x }{ 5 }\) – \(\frac { 2y }{ 3 }\) + 1 = 0, contains the point (m, 2m – 1); calculate the value of m.
Solution:
Point (m, 2m -1) satisfies the equation

Question 6.
Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2) ?
Solution:
Line 3x – 5y = 6 bisect the join of points (5, -2) and (-1, 2)
The mid-point of (5, -2) and (-1, 2) satisfies the equation.
Now, mid-point of (5, -2) and (-1, 2)

Now, substituting x = 2, y = 0, in the given equation
3 x 2 – 5 x 0 = 6 ⇒ 6 – 0 = 6 ⇒ 6 = 6 which is true. .
Given line bisect the join of points (5, -2) and (-1, 2)

Question 7.
(i) The line y = 3x – 2 bisects the join of (a, 3) and (2, -5), find the value of k.
(ii) The line x – 6y + 11 = 0 bisects the join of (8, -1) and ( 0, k). Find the value of k.
Solution:
(i) line y = 3x – 2 bisects the join of (a, 3) and (2, -5)
Mid-point join of there points satisfies it.
Now, mid-point of (a, 3) and (2, -5) is

Question 8.
(i) The point (-3, -2) lies on the line ax + 3y + 6 = 0, calculate the value of ‘a’
(ii) The line y = mx + 8 contains the point (- 4, 4), calculate the value of ‘m’
Solution:
(i) Point (-3, 2) lies on the line ax + 3y + 6 = 0,
Then x = – 3, y = 2 satisfies it
a (-3) + 3(2) + 6 = 0
⇒ -3a + 6 + 6 = 0
⇒ -3a + 12 = 0
⇒ -3a = – 12
⇒ a = 4
(ii) line y = mx + 8 contains the point (-4, 4)
x = – 4, y = 4 satisfies it
4 = m (-4) + 8
⇒ 4 = -4m + 8
⇒ 4m = 8 – 4 = 4
⇒ m = 1

Question 9.
The point P divides the join of (2, 1) and (-3, 6) in the ratio 2 : 3. Does P lie on the line x – 5y + 15 = 0 ?
Solution:
P divides the line joining of the points (2, 1) and (-3, 6) in the ratio of 2 : 3,
co-ordinates of P will be

Now, substituting x = 0, y = 3 in the equation
x – 5y + 15 = 0
⇒ 0 – 5 x 3 + 15 = 0
⇒ 0 – 15 + 15 = 0
⇒ 0 = 0 which is true.
Point (0, 3) lies on the line.

Question 10.
The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio of 1 : 2. Does the line x – 2y = 0 contain Q ?
Solution:
Point Q, divides the line segment joining the points (5, -4) and (2, 2) in the rates of 1 : 2
co-ordinates of Q will be,

Now, substituting x = 4, y = – 2 in the equation
x – 2y = 0, we get
4 – 2 x (-2) = 0
⇒ 4 + 4 = 0
⇒ 8 = 0 which is not true.
Point Q does not lie on the line x – 2y = 0

Question 11.
Find the point of intersection of the lines : 4x + 3y = 1 and 3x – y + 9 = 0. If this point lies on the line (2k – 1) x – 2y = 4; find the value of k.
Solution:
4x + 3y = 1 …..(i)
3x – y = -9 …..(ii).
Multiplying (i) by 1 and (ii) 3
4x + 3y = 1
9x – 3y = -27
Adding, we get-:
1 3x = – 26 ⇒ x = -2
from (ii),
3x – y = – 9
3(-2) – y = -9
⇒ – 6 – y = -9
⇒ -y = -9 + 6 = -3
⇒ y = 3
The point of intersection is (-2, 3)
The line (2k – 1) x – 2y = 4 passes through that point also
It is satisfy it.
(2k – 1) (-2) – 2(3) = 4
⇒ -4k + 2 – 6 = 4
⇒ -4k – 4 = 4
⇒ -4k = 4 + 4 = 8
⇒ k = -2
Hence point of intersection is (-2, 3) and value of k = -2

Question 12.
Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent.
Solution:
2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent
They will pass through the same point
Now 2x + 5y = 1 …..(i)
x – 3y = 6 ……(ii)
Multiply (i) by 3 and (ii) by 5, we get :
-6x + 15y = 3
5x – 15y = 30
Adding we get :
11x = 33 ⇒ x = 3
from (ii),
x – 3y = 6
⇒ 3 – 3y = 6
⇒ -3y = 6 – 3 = 3
⇒ y = -1
Point of intersection of first two lines is (3, -1)
Substituting the values in third line x + 5y + 2 = 0
L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 3 – 5 + 2 = 5 – 5 = 0 = R.H.S.
Hence the given three lines are concurrent.

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

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C

Question 1.
Given a triangle ABC in which A = (4, -4), B (0, 5) and C = (5, 10). A point P lies on BC such that BP : PC = 3 : 2. Find the length of line segment AP.
Solution:
B (0, 5), C (5, 10) and BP : PC = 3 : 2 Co-ordinates of P will be

Question 2.
A (20, 0) and B (10, – 20) are two fixed points, find the co-ordinates of the point P in AB such that 3PB = AB. Also, find the co-ordinates of some other point Q in AB such that AB = 6AQ.
Solution:
(i) A (20, 0), B (10, – 20)

Question 3.
A (-8, 0), B (0, 16) and C (0, 0) are the vertices of a triangle ABC. Point P lies on AB and Q lies on AC such that AP : PB = 3 : 5 and AQ : QC = 3 : 5. Show that: PQ = \(\frac { 3 }{ 8 }\) BC.
Solution:

Question 4.
Find the co-ordinates of points of trisection of the line segment joining the points (6, -9) and the origin.
Solution:
Points are A (6, -9) and O (0,0) let P and Q are points, which trisect AO

Question 5.
A line segment joining A (-1, \(\frac { 5 }{ 3 }\)) and B (a, 5) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects the y-axis.
(i) Calculate the value of ‘a’.
(ii) Calculate the co-ordinates of ‘P’. (1994)
Solution:

Question 6.
In what ratio is the line joining A (0, 3) and B (4, -1), divided by the x-axis ? Write the co-ordinates of the point where AB intersects the x-axis. [1993]
Solution:
Let the ratio be m₁ : m₂ when the x-axis intersects the line AB at P.

Question 7.
The mid point of the segment AB, as shown in diagram, is C (4, -3). Write down the co-ordinates of A and B. (1996)

Solution:
Let co-ordinates of A (x, 0) and B (0, y) and C (4, -3) the mid point of AB.

Question 8.
AB is a diameter of a circle with centre C = (-2, 5). If A = (3, -7). Find
(i) the length of radius AC
(ii) the coordinates of B.
Solution:

Question 9.
Find the co-ordinates of the centroid of a triangle ABC whose vertices are A (- 1, 3), B (1, – 1) and C (5, 1)
Solution:

Question 10.
The mid-point of the line segment joining (4a, 2b – 3) and (-4, 3b) is (2, -2a). Find the values of a and b.
Solution:
Let A and B are two points and P is its mid point then A is (4a, 2b -3), B(-4, 2b) and P is (2, -2a)

Question 11.
The mid point of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1). Find the value of a and b.
Solution:
The midpoint of the line segment joining (2a, 4) and (-2, 2b) is (1, 2a + 1)

Question 12.
(i) Write down the co-ordinates of the point P that divides the line joining A (-4, 1) and B (17, 10) in the ratio 1 : 2.
(ii) Calculate the distance OP, where O is the origin.
(iii) In what ratio does the y-axis divide the line AB ? [ICSE 1995]
Solution:
Point P, divides a line segment giving the points A (-4, 1) and B (17, 10) is the ratio 1 : 2.

Question 13.
Prove that the points A(-5, 4); B (-1, -2) and C (5, 2) are the vertices of an isosceles right-angled triangle. Find the co-ordinates of D. So that ABCD is a square. [1992]
Solution:
In ABC, the co-ordinates of A, B and C are (-5, 4), B(-1, -2) and C (5, 2) respectively.

ABC is also a right-angled triangle.
Hence ABC is an isosceles right angled triangle,
Let D be the fourth vertex of square ABCD and co-ordinates of D be (x,y)
Since the diagonals of a square bisect each other and let O be the point of intersection of AC and BD.
O is mid-point of AC as well as BD.

Question 14.
M is the mid-point of the line segment joining the points A (-3, 7) and B (9, -1). Find the co-ordinates of point M. Further, if R (2, 2) divides the line segment joining M and the origin in the ratio p : q, find the ratio p : q.
Solution:
Two points are given A (-3, 7) and B (9, -1)
M is the mid-point of line joining AB.
Co-ordinates of M wll be

Question 15.
Calculate the ratio in which the line joining A (-4, 2) and B (3, 6) is divided by point P (x, 3). Also find
(i) x
(ii) Length of AP. (2014)
Solution:
Let ratio = k : 1

Question 16.
Find the ratio in which the line 2x + y = 4 divides the line segment joining the points P (2, -2) and Q (3, 7).
Solution:
Let the given line 2x + y = 4 divides the line segment joining the points P (2, -2) and Q (3,7) in the ratio k : 1 at a point (x, y) on it.

Question 17.
If the abscissa of a point P is 2. Find the ratio in which this point divides the line segment joining the points (-4, 3) and (6, 3). Also, find the co-ordinate of point P.
Solution:
Abscissa of a point P is 2
Let co-ordinates of point P be (2, y)
Let point P (2, y) divides the line segment joining the points (-4, 3) and (6, 3) in the ratio k : 1

Question 18.
The line joining the points (2, 1) and (5, -8) is trisected at the points P and Q. If point P lies on the line 2x – y + k = 0, find the value of k, Also, find the co-ordinates of point Q.
Solution:
A line joining the points (2, 1) and (5, -8) is trisector at P and Q.

Question 19.
M is the mid-point of the line segment joining the points A (0, 4) and B (6, 0). M also divides the line segment OP in the ratio 1 : 3. Find:
(i) co-ordinates of M
(ii) co-ordinates of P
(iii) length of BP

Solution:
M is mid point of the line segment joining the points A (0, 4) and B (6, 0)
M divides the line segment OP in the ratio 1 : 3

Question 20.
Find the image of the point A (5, -3) under reflection in the point P (-1, 3).
Solution:
Image of the point A (5, -3) under reflection in the point P (-1, 3)
Let B (x, y) be the point of reflection of A (5, -3) under P(-1, 3)

Question 21.
A (-4, 2), B (0, 2) and C (-2, -4) are vertices of a triangle ABC. P, Q and R are mid-points of sides BC, CA and AB respectively. Show that the centroid of PQR is the same as the centroid of ABC.
Solution:
A (-4, 2), B (0, 2) and C (-2, -4) are the vertices of ABC.
P, Q and R are the mid-points of the sides BC, CA and AB respectively.
G is the centroid of medians AP, BQ and CR.
Co-ordinates of G are

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C 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 13 Section and Mid-Point Formula Ex 13B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C

Question 1.
Find the mid-point of the line segment joining the points:
(i) (-6, 7) and (3, 5)
(ii) (5, -3), (-1, 7)
Solution:
Let P (x, y) be the mid-point in each case

Question 2.
Points A and B have co-ordinates (3, 5) and (x, y) respectively. The mid-point of AB is (2, 3). Find the values of x and y.
Solution:
Co-ordinates of A (3, 5), B (x, y) and mid-point M (2, 3)

Question 3.
A (5, 3), B (-1, 1) and C (7, -3) are the vertices of ABC. If L is the mid-point of AB and M is the mid-point of AC, show that LM = \(\frac { 1 }{ 2 }\) BC.
Solution:

Question 4.
Given M is the mid-point of AB, find the co-ordinates of:
(i) A; if M = (1, 7) and B = (-5, 10),
(ii) B; if A = (3, -1) and M (-1, 3).
Solution:
M is the mid-point of AB.
(i) Let A = (x, y), M = (1, 7) and B = (-5, 10)

Question 5.
P (-3, 2) is the mid-point of line segment AB as shown in the figure. Find the co-ordinates of points A and B.

Solution:

Point A is on y-axis
its abscissa is zero and point B is on x-axis
its ordinate is zero.
Now, let co-ordinates of A are (0, y) and ofB are (x, 0) and P (-3, 2) is the mid-point

Question 6.
In the given figure, P (4, 2) is the mid point of line segment AB. Find the co-ordinates of A and B.

Solution:

Points A and B are on x-axis and y-axis respectively
Ordinate of A is zero and abscissa of B is zero.
Let co-ordinates of A be (x, 0) and B (0, y)
and P (4, 2) is the mid-point

Question 7.
(-5, 2), (3, -6) and (7, 4) arc the vertices of a triangle. Find the length of its median through the vertex (3, -6) and (7, 4).
Solution:

Let A (-5, 2), B (3, -6) and C (7, 4) are the vertices of a ABC
Let L,M and N are the mid-points of sides BC, CA and AB respectively of ABC.
L is the mid-point of BC.
Co-ordinates of L will be

Question 8.
Given a line ABCD in which AB = BC = CD, B = (0, 3) and C = (1, 8). Find the co-ordinates of A and D.

Solution:

Question 9.
One end of the diameter of a circle is (-2, 5). Find the co-ordinates of the other end of it, if the centre of the circle is (2, -1).
Solution:

Question 10.
A (2, 5), B (1, 0), C (-4, 3) and D (-3, 8) are the vertices of quadrilateral ABCD. Find the co-ordinates of the mid-points of AC and BD. Give a special name to the quadrilateral.
Solution:
Co-ordinates of A = (2, 5), B = (1, 0), C = (-4, 3) and D = ( 3, 8)

Let the mid-point of AC is P (x1, y1) Co-ordinates of mid-point of AC will be

Co-ordinates of mid-points AC and BD are the same..
The quadrilateral is a parallelogram.

Question 11.
P (4, 2) and Q (-1, 5) are the vertices of parallelogram PQRS and (-3, 2) are the co-ordinates of the point of intersection of its diagonals. Find the co-ordinates of R and S.
Solution:
In the parallelogram PQRS and qo-ordinates of P are (4, 2) and of Q are (-1, 5).
The diagonals of || gm AC and BD intersect each other at O (-3, 2)

Question 12.
A (-1, 0), B (1, 3) and D (3, 5) are the vertices of a parallelogram ABCD. Find the co-ordinates of vertex C.
Solution:

Vertices of a parallelogram ABCD are A (-1, 0), B (1, 3) and D(3, 5)
Let co-ordinates of C be (x, y)
Let the diagonals AC and BD bisect each other at O. Then O is the mid-point of AC as well as of BD.
Co-ordinates of O, the mid-point of BD will be

Question 13.
The points (2, -1), (-1, 4) and (-2, 2) are the mid-points of the sides of a triangle. Find its vertices.
Solution:
Let D, E and F are the mid-points of sides BC, CA and AB of a ABC respectively.


Co-ordinates of A are (-5, 7), of B are (1, -3) and of C are (3, 1)

Question 14.
Points A (-5, x), B (y, 7) and C (1, -3) are collinear (i.e. lie on the same straight line) such that AB = BC. Calculate the values of x and y.
Solution:

Question 15.
Points P (a, -4), Q (-2, b) and R (0, 2) are collinear. If Q lies between P and R, such that PR = 2QR, calculate the values of ‘a’ and ‘b’:
Solution:

Question 16.
Calculate the co-ordinates of the centroid of the triangle ABC, if A = (7, -2), B = (0, 1) and C = (-1, 4).
Solution:

Question 17.
The co-ordinates of the centroid of a triangle PQR are (2, -5). If Q = (-6, 5) and R = (11, 8); calculate the co-ordinates of vertex P.
Solution:

Question 18.
A (5, x), B (-4, 3) and C (y, -2) are the vertices of the triangle ABC whose centroid is the origin. Calculate the values of x and y.
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B 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 13 Section and Mid-Point Formula Ex 13A

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13C

Question 1.
Calculate the co-ordinates-of the point P which divides the line segment joining:
(i) A (1, 3) and B (5, 9) in the ratio 1 : 2
(ii) A (-4, 6) and B (3, -5) in the ratio 3 : 2
Solution:
(i) Let co-ordinates of P be (x,y)

Question 2.
In what ratio is the line joining (2, -3) and (5, 6) divided by the x-axis ?
Solution:
Let the point P (x, 0) divides in the ratio of m₁ : m₂ line joining the points A (2, -3) and B (5, 6)

Question 3.
In what ratio is the line joining (2, -4) and (-3, 6) divided by the y-axis ?
Solution:
Let the point P (0, y) divides the line joining the points A (2, -4) and (-3, 6) in the ratio of m₁ : m₂

Question 4.
In what ratio does the point (1, a) divide the join of (-1, 4) and (4, -1)? Also, find the value of ‘a’.
Solution:
Let the point P (1, a) divides the line joining the points (-1, 4) and (4, -1) in the ratio of m₁ : m₂

Question 5.
In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8) ? Also, find the value of ‘a’.
Solution:
Let the point P (a, 6) divides the line joining the points A (-4, 3), B (2, 8) in the ratio of m₁ : m₂

Question 6.
In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.
Solution:

Question 7.
Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the co-ordinates of the point of intersection.
Solution:
Let, the points (0, y) be the point of intersection which divides the line joining the points A (-4, 7) and B (3, 0)

Question 8.
Points A, B, C and D divide the line segment joining the point (5, -10) and the origin in five equal parts. Find the co-ordinates of A, B, C and D.
Solution:
Points A, B, C and D divide the line segment joining the points (5, -10) and origin (0, 0) in five equal parts
Let co-ordinates of A be (x, y) which divides PO in the ratio of 1 : 4

Question 9.
The line joining the points A (-3, -10) and B (-2, 6) is divided by the point P such that = \(\frac { PA }{ PB }\) = \(\frac { 1 }{ 5 }\), find the co-ordinates of P.
Solution:
Let the co-ordinates of P be (x, y) which divides the line joining the points A (-3,-10) and B (-2,6) in the ratio of AP : PB i.e. (5 – 1) : 1 or 4 : 1

Question 10.
P is a point on the line joining A (4, 3) and B (-2, 6) such that 5AP = 2BP. Find the co-ordinates of P.
Solution:

Question 11.
Calculate the ratio in which the line joining the points (-3, -1) and (5, 7) is divided by the line x = 2. Also, find the co-ordinates of the point of intersection.
Solution:
Let the point P (2, y) divides the line joining the points A (-3, -1) and B (5, 7) in the ratio of m₁ : m₂

Question 12.
Calculate the ratio in which the line joining A (6, 5) and B (4, -3) is divided by the line y = 2.
Solution:
Let the point P (x, 2) divides the line joining the points A (6, 5) and B (4, -3) in the ratio of m₁ : m₂

Question 13.
The point P(5, -4) divides the line segment AB, as shown in the figure, in the ratio 2 : 5. Find the co-ordinates of points A and B.

Solution:
From the figure, the line AB intersects x-axis at A and y-axis at B.
Let the co-ordinates of A (x, 0) and B (0, y) and P (5, -4) divides it in the ratio of 2 : 5

Question 14.
Find the co-ordinates of the points of trisection of the line joining the points (-3, 0) and (6, 6).
Solution:

Question 15.
Show that the line segment joining the points (-5, 8) and (10, -4) is trisected by the co-ordinate axes.
Solution:

Let the points A (-5, 8) and B (10, -4).
Let P and Q be the two points on the axis which trisect the line joining the points A and B.
AP = PQ = QB
AP : PB = 1 : 2 and AQ : QB = 2 : 1

Question 16.
Show that A (3, -2) is a point of trisection of the line-segment joining the points (2, 1) and (5, -8). Also, find the co-ordinates of the other point of trisection.
Solution:

Let A and B are the points of trisection of the line segment joining the points P (2, 1) and Q (5, -8), then
PA = AB = BQ.
PA : AQ = 1 : 2 and PB : BQ = 2 : 1

Question 17.
If A = (-4, 3) and B = (8, -6)
(i) find the length of AB
(ii) In what ratio is the line joining A and B, divided by the x-axis ?
Solution:

Question 18.
The line segment joining the points M (5, 7) and N (-3, 2) is intersected by the y-axis at point L. Write down the abscissa of L. Hence, find the ratio in which L divides MN. Also, find the co-ordinates of L.
Solution:

Question 19.
A (2, 5), B (-1, 2) and C (5, 8) are the co-ordinates of the vertices of the triangle ABC. Points P and Q lie on AB and AC respectively, such that:
AP : PB = AQ : QC = 1 : 2.
(i) Calculate the co-ordinates of P and Q.
(ii) Show that PQ = \(\frac { 1 }{ 3 }\) BC.
Solution:

Question 20.
A (-3, 4), B ( 3, -1) and C (-2, 4) are the vertices of a triangle ABC. Find the length of line segment AP, where point P lies inside BC, such that BP: PC = 2 : 3.
Solution:

Question 21.
ThelinesegmentjoiningA(2, 3)andB(6, -5) is intercepted by x-axis at the point K. Write down the ordinate of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the pointK. [1990, 2006]
Solution:
Let the line segment Intersect the x-axis at the point P
Co-ordinates of P are (x, 0)
Let P divide the line segment in the ratio K : 1 then

Question 22.
The line segment joining A (4, 7) and B (-6, -2) is intercepted by the y-axis at the point K. Write down the abscissa of the point K. Hence, find the ratio in which K divides AB. Also, find the co-ordinates of the point K.
Solution:
Points A (4, -7), B (-6, -2) are joined which intersects y-axis at K. abscissa of K will be 0

Let the coordinates of K be (0, y) and K divides AB line segment in the ratio m₁ : m₂

Question 23.
The line joining P (-4, 5) and Q (3, 2), intersects they axis at point R. PM and QN are perpendiculars from P and Q on the x-axis. Find:
(i) The ratio PR: RQ.
(ii) The co-ordinates of R.
(iii) The areas of the quadrilateral PMNQ. [2004]
Solution:
(i) Let divides the line joining the points P (-4, 5) and Q (3, 2) in the ratio k : 1

Question 24.
In the given figure, line APB meets the x- axis at point A and y-axis at point B. P is the point (-4, 2) and AP : PB = 1 : 2. Find the co-ordinates of A and B.

Solution:
Let the co-ordinates of A be (x1, 0) (as it lies on x-axis)
and co-ordinates of B be (0, y2)
and co-ordinates of P are (-4, 2)
AP : PB = 1 : 2 i.e. m1 = 1, m2 = 2
Now, P divides AB in the ratio m₁ : m₂ or 1 : 2

Question 25.
Given a line segment AB joining the points A (-4, 6) and B (8, -3). Find:
(i) the ratio in which AB is divided by the y-axis.
(ii) find the coordinates of the point of intersection.
(iii) the length of AB.
Solution:
(i) Let the y-axis divide AB in the ratio m : 1

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 13 Section and Mid-Point Formula Ex 13A 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 12 Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points) Ex 12B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 12 Reflection Ex 12B.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 12 Reflection Ex 12A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 12 Reflection Ex 12B

Question 1.
Attempt this question on graph paper.
(a) Plot A (3, 2) and B (5, 4) on graph paper. Take 2cm = 1 unit on both the axes.
(b) Reflect A and B in the x-axis to A’ and B’ respectively. Plot these points also on the same graph paper.
(c) Write down:
(i) The geometrical name of the figure ABB’A’;
(ii) The measure of angle ABB’ ;
(iii) The image A” of A, when A is reflected in the origin.
(d) The single transformation that maps A’ to A”. [1995]
Solution:
From the graph, we can say that
(iii) (a) an isosceles trapezium
(b) 45°
(c) ( -3, -2)
(d) reflection in y-axis

Question 2.
Points (3,0) and (-1,0) are invariant points under reflection in the line L1; points (0, -3) and (0, 1) are invariant points on reflection in line L2
(a) Name or write equations for the lines L1 and L2 .
(b) Write down the images of points P (3, 4) and Q ( -5, -2) on reflection in L. Name the images as P’ and Q’ respectively.
(c) Write down the images of P and Q on re-flection in L,. Name the images as P” and Q” respectively.
(d) State or describe a single transformation that maps P’ onto P”. [1996]
Solution:

(a) Points (3, 0) and (-1, 0) are invariant points under reflection L .
Here the ordinates of both points are 0 It is x-axis or y = 0.
Again points (0, -3) and (0, 1) are invariant- points on reflection in line L2
Here the abscissas are 0 in both points.
It is y-axis or x = 0.
(b) The co-ordinates of the images of points P (3, 4) and Q (-5, -2) on reflection in L, will be P (3, -4) and Q'(-5, 2)
(c) The co-ordinate of images of points P (3, 4) and Q (-5, -2) on reflection in L2 will be P’ (-3, 4) and Q’ (5, -2)
(d) Reflection is in origin because the co-ordinates of P’ and P” have opposite signs.

Question 3.
(a) Point P (a, b) is reflected in the x-axis to P’ (5, -2), write down the values of a and b.
(b) P” is the image of P when reflected in y- axis, write down the co-ordinates of P”.
(c) Name a single transformation that maps P’ to P”. (1997):
Solution:
(a) P (a, b) is reflected in x-axis to P’ (5, -2). the co-ordinates of P will be (5, 2)
a = 5, b = 2
(b) P” is the image of P when reflected is y-axis.
Co-ordinates of P” will be ( -5, 2)
(c) Reflection is in origin.

Question 4.
The point (-2, 0) on reflection in a line is mapped to (2, 0) and the point (5, -6) on reflection in the same line is mapped to (-5, -6).
(a) State the name of the mirror line and write its equation.
(b) State the co-ordinates of the image of (-8, -5) in the mirror line.
Solution:
Point (-2, 0) is mapped to (2, 0) to a certain line and point (5, -6) is also mapped to (-5, -6) to the same line.
(a) We see that sign of x-coordinate is changed.
The required line is y-axis whose equation will be x = 0.
(b) The co-ordinates of the image of the point (-8, -5) in the same mirror line will be (8, -5).

Question 5.
The points P (4, 1) and Q (-2, 4) are reflected in line y = 3, Find the co-ordinates of P’, the image of P and Q’, the image of Q.
Solution:
Co-ordinates of P and Q are (4, 1) and ( -2, 4) respectively.

The co-ordinates of image of P which is P’ are (4, 5) reflection in the line y = 3
and co-ordinates of image of Q when is Q’ are ( -2, 2) reflection is the line y = 3 as shown in the graph.

Question 6.
A point P (-2, 3) is reflected in the line x = 2 to point P’. Find the co-ordinates of P’.
Solution:
The image of P ( -2, 3) is P’ which is reflected in the line x = 2. The co-ordinates of P’ will be (6, 3) as shown in the graph.

Question 7.
A point P (a, b) is reflected in the x-axis to P’ (2, -3). Write down the values of a and b. P” is the image of P, reflected in the y- axis. Write down the co-ordinates of P”. Find the co-ordinates of P'”, when P is reflected in the line, parallel to y-axis, such that x = 4. [1998]
Solution:

(i) A point P (a, b) is reflected in v-axis to P’ (2, -3)
Co-ordinates of P will be (2, 3) a = 2, b = 3.
(ii) P” is the image of P, when reflected in the y- axis
Co-ordinates of P” will be ( -2, 3)
(iii) The image of P is P'” reflected in a line parallel to y-axis i.e. x = 4.
Co-ordinates of P”‘ will be (6, 3)

Question 8.
Points A and B have co-ordinates (3, 4) and (0, 2) respectively. Find the image:
(a) A’ of A under reflection in the x-axis.
(b) B’ of B under reflection in the line AA’.
(c) A” of A under reflection in the y-axis.
(d) B” of B under reflection in the line AA”.
Solution:
Co-ordinates of A are (3, 4) and co-ordinates of B are (0, 2)
(a) Co-ordinates, of A’ of A under reflection in the v-axis are (3, -4)
(b) Co-ordinates of B’ of B under reflection in the line AA’ are (6, 2)
(c) Co-ordinates of A” of A under reflection in y-axis are (-3, 4)
(d) Co-ordinates of B” of B under reflection in the line A A” are (0, 8).

Question 9.
(a) Plot the points A(3, 5) and B(-2, -4). Use 1 cm = 1 unit on both the axes.
(b) A’ is the image of A when reflected in the v-axis. Write down the co-ordinates of A’ and plot it on the graph paper.
(c) B’ is the image of B when reflected in the y-axis, followed by reflection in the origin.
Write down the co-ordinates of B’ and plot it on the graph paper.
(d) Write down the geometrical name of the figure AA’ BB’.
(e) Name two invariant points under reflection in the x-axis. [1999]
Solution:
(a) The points A (3, 5) and B(-2, -4) have been plotted on the graph.
(b) A’ is the image of A when reflected in the x-axis

Co-ordinates of A’ will be (3, -5)
(c) B’ is the image of B when reflected in y- axis, followed by reflection in the origin.
Co-ordinates will be (-2, 4)
Images A’ and B’ have also been plotted.
(d) Join AA’, A’B, BB’ and B’A the quadrilateral so joined is an isosceles trapezium.
(e) The points C and D whose co-ordinates ! are (3,0) and (-2, 0) are invariant points under reflection in the x-axis.

Question 10.
The point P (5,3) was reflected in the origin to get the image P’.
(a) Write down the co-ordinates of P’.
(b) If M is the foot of the perpendicular from P to the x-axis, find the co-ordinates of M.
(c) If N is the foot of the perpendicular from P’ to the x-axis, find the co-ordinates of N.
(d) Name the figure PM P’ N.
(e) Find the area of the figure PM P’ N.
Solution:
(a) Points P(5, 3) is reflected to P’ in the origin.
Co-ordinates of P’ will be (-5, -3).
(b) PM ⊥ on x-axis.
Co-ordinates of P’ will be (5, 0)
(c) P’N X on x-axis
Coordinates of N will be (-5, 0).
(d) PM, MP’ P’N and N. P are joined which form a parallelogram PMP’N.
(e) Area of || gm PMP’N = 2 (Area of ∆PMN)
= 2 (\(\frac { 1 }{ 2 }\) x 10 x 3) Sq. cm
= 30 Sq. cm

Question 11.
The point P (3, 4) is reflected to P’ in the x-axis; and O’ is the image of O (the origin) when reflected in the line PP’. Write :
(i) the co-ordinates of P’ and O’,
(ii) the length of the segments PP’ and OO’,
(iii) the perimeter of the quadrilateral POP’O’.
(iv) the geometrical name of the figure POP’O’.
Solution:

(i) Point P (3, 4) is reflected to P’ in x-axis and then its co-ordinates will be (3, -4) again O’ is the image of O (0, 0) reflected in PP’, then co-ordinates of O’ will be (6, 0)
(ii) Length of PP’ = 4 + 4 = 8 units and OO’ = 3 + 3 = 6 units
(iii) Diagonals of OPO’P’ bisect each other at right angles at M.
OPO’P’ is a rhombus.
(iv) The perimeter of quadrilateral OPO’P’ = 5 x 4 = 20 units
(v) The quadrilateral OPO’P’ is a rhombus as rhombus shown on the graph paper.

Question 12.
A (1, 1), B (5, 1), C (4, 2) and D (2, 2) are vertices of a quadrilateral. Name the quadrilateral ABCD. A, B, C, and D are reflected in the origin on to A’, B’, C’ and D’ respectively. Locate A’, B’, C’ and D’ on the graph sheet and write their co-ordinates. Are D, A, A’ and D’ collinear? (2004)
Solution:
On the given graph, plot the points A(1, 1), B(5, 1). C(4, 2) and D(2, 2) Join AB, BC, CD and DA.

The quadrilateral is of trapezium in shape.
Now points A, B, C and D are reflected in the origin on to A’, B’, C and D’ respectively.
Co-ordinates of A’ are (-1, -1), of B’ are (-5, -1), of C are (-4, -2) and of D’ are (-2, -2).
A’ B, B’C’, CD’ and D’A’ are joined.
Yes, we see that DA A’D’ are collinear.

Question 13.
P and Q have co-ordinates (0, 5) and (-2, 4).
(a) P is invariant when reflected in an axis, Name the axis.
(b) Find the image of Q on reflection in the axis found in fa)
(c) (0, k) on reflection in the origin is invariant. Write the value of k.
(d) Write the co-ordinates of the image of Q, obtained by reflecting it in the origin followed by reflection in x-axis. (2005)
Solution:
Co-ordinates of the given points P and Q, are (0, 5) and (-2, 4) and have been plotted on the graph.

(a) P is invariant in y-axis
(b) Image of Q(-2, 4) is Q’ in years and its Co-ordinates will be (2, 4)
(c) (0, k) on reflections in we origin is also invariant
Co-ordinates will be (0, 0) k = 0
(d) Let Q” be image of Q reflection in his origin.
Again Q” is reflected in x-axis and its image will be Q’ (2, 4)

Question 14.
(a) The point P (2, -4) is reflected about the line x = 0 to get the image Q. Find the co-ordinates of Q.
(b) The point Q is reflected about the line y = 0 to get the image R. Find the co-ordinates of R.
(c) Name the figure PQR.
(d) Find the area of figure PQR.
Solution:


The graph of the solution set is shown by the thick portion of the number line. The solid circle at -3 indicates that the number – 3 is included among the solutions whereas the open circle at 3 indicates that 3 is not included among the solutions.
(c) (i) Since the point Q is the reflection of the point P (2, – 4) in the line x = 0, the co-ordinates of Q are (2, 4).
(ii) Since R is the reflection of Q (2, 4) about the line x = 0, the co-ordinates of R are (- 2, 4).
(iii) Figure PQR is the right angled triangle PQR.
(d) Area of ∆PQR = \(\frac { 1 }{ 2 }\) x QR x PQ = \(\frac { 1 }{ 2 }\) x 4 x 8 = 16 sq. units.

Question 15.
Using a graph paper, plot the points A (6, 4) and B (0, 4).
(i) Reflect A and B in the origin to get the images A’ and B’.
(ii) Write the co-ordinates of A’ and B’.
(iii) State the geometrical name for the figure ABA’ B’
(iv) Find its perimeter.
Solution:

Question 16.
Use graph paper for this question. (Take 2 cm = 1 unit along both x and y axis.
Plot the points O (0, 0), A (-4, 4), B (-3, 0) and C (0, -3)
(i) Reflect points A and B on the y-axis and name them A’ and B’ respectively. Write down their coordinates. k
(ii) Name the figure OABCB’A’.
(iii) State the line of symmetry of this figure. (2016)
Solution:


(ii) The figure OABCB’A’ is similar to arrow headed.
(iii) The y-axis is the line of symmetry of figure OABCB’A’.

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 12 Reflection Ex 12B 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.