## 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**

- 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 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

**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:**

**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:**

**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

**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

**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

**Question 10.**

**If the lines y = 3x + 7 and 2y +px = 3 are perpendicular to each other, find the value of p. (2006)**

**Solution:**

**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)

**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

⇒ 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

**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)

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.**

**Solution:**

(i) From the figure, the see that co-ordinates of A are (2, 3), B are (-1, 2) and C are (3, 0)

**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:**

**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:**

**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

**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:

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

**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)

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:**

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]**

**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

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