# ICSE Solutions for Chapter 14 Equation of a Line Class 10 Mathematics

**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 1:**Using section formula, the co-ordinates of the point P are

(3×16 + 5×8)/(3 + 5), (3×(-8) + 5×0)/(3 + 5)}

= (11, -3) = (x

_{1}, y

_{1})

3x + 5y = 7

⇒ y = -3/5x + 7/5

Slope of this line = -3/5

As the required line is parallel to the line 3x + 5y = 7.

Slope of the required line = Slope of the given line = -3/5

Thus, the equation of the required line is

y – y

_{1}= m(x – x

_{1})

y + 3 = -3/5(x – 11)

5y + 15 = - 3x + 33

3x + 5y = 18

**Question 2: The equation of a line x - y = 4. Find its slope and y-intercept. Also, find its inclination.**

**Solution 2:**Given equation of a line is x – y = 4

⇒ y = x - 4

Comparing this equation with y = mx + c. We have:

Slope = m = 1

y-intercept = c = -4

Let the inclination be Î¸.

Slope = 1 = tanÎ¸ = tan 45˚

∴ Î¸ = 45°

**Question 3: Find the equation of the line whose slope is -4/3**

**and which passes through (-3, 4).**

**Solution 3:**Given, slopes= -4/3

The equation passes through (-3, 4) (x

_{1}, y

_{1})

Substituting the values in y – y

_{1}= m(x – x

_{1}), we get

y – 4 = -4/3(x + 3)

3y - 12 = -4x - 12

4x + 3y = 0, which is the required equation.

**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 4:**

(i) Slope of AB = (6 – 4)/(0 + 2) = 2/2 = 1

Slope of the line parallel to AB = Slope of AB = 1

(ii) Slope of AB = (5 + 3)/(-2 – 0) = 8/-2 = -4

Slope of the line parallel to AB = Slope of AB = -4

**Question 5: The line 3x/5 – 2y/3 + 1 = 0 contains the point (m, 2m - 1); calculate the value of m.**

**Solution 5:**

The equation of the given line is 3x/5 – 2y/3 + 1 = 0

Putting x = m, y = 2m - 1, we have:

3m/5 – 2(2m – 1)/3 + 1 = 0

3m/5 – (4m – 2)/3 = -1

(9m – 20m + 10)/15 = -1

9m – 20m + 10 = -15

-11m = -25

m = 25/11 = 2 3/11

**Question 6: Does the line 3x - 5y = 6 bisect the join of (5, -2) and (-1, 2)?**

**Solution 6:**The given line will bisect the join of A (5, -2) and B (-1, 2), if the co-ordinates of the mid-point of AB satisfy the equation of the line.

The co-ordinates of the mid-point of AB are

{(5 – 1)/2, (-2 + 2)/2} = (2, 0)

Substituting x = 2 and y = 0 in the given equation, we have:

L.H.S. = 3x - 5y = 3(2) - 5(0) = 6 – 0 = 6 = R.H.S.

Hence, the line 3x - 5y = 6 bisect the join of (5, -2) and (-1, 2).

**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 7:**

Slope of the line passing through (-4, -2) and (2, -3) = (-3 + 2)/(2 + 4) = -1/6

Slope of the line passing through (a, 5) and (2, -1) = (-1 – 5)/(2 – a) = -6/(2 – a)

Since, the lines are perpendicular.

∴ -1/6 = -1/{-6/(2 – a)}

-1/6 = (2 – a)/6

2 – a = -1

a = 3

**Question 8: The figure given below shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equation of AB and CD.**

**Solution 8:**Slope of line AB = tan 45˚ = 1

The line AB passes through P (3, 4). So, the equation of the line AB is given by:

y – y

_{1}= m(x – x

_{1})

y – 4 = 1(x – 3)

y – 4 = x – 3

y = x + 1

Slope of line CD = tan 60° = √3

The line CD passes through P (3,4). So, the equation of the line CD is given by:

y - y

_{1 }m(x - x

_{1})

y – 4 = √3(x - 3)

y – 4 = √3× -3√3

y = √3x + 4 - 3√3

**Question 9: The lines represented by 4x + 3y = 9 and px - 6y + 3 = 0 are parallel. Find the value of p.**

**Solution 9:**

4x + 3y = 9

3y = -4x + 9

y = -4/3x + 3

Slope of this line = -4/3

px - 6y + 3 = 0

6y = px + 3

y = P/6x + 1/2

Slope of this line = P/6

Since the lines are parallel, their slopes will be equal.

∴ -4/3 = P/6

-4 = P/2

P = - 8

**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 10:**7x + 6y = 71

⇒ 28x + 24 = 284

5x - 8y = -23

**...(1)**5x - 8y = -23

⇒ 15x - 24y = -69

Adding (1) and (2), we get,

43x = 215

⇒ x = 5

From (2), 8y = 5x + 23

**…(2)**Adding (1) and (2), we get,

43x = 215

⇒ x = 5

From (2), 8y = 5x + 23

Putting value of x,

25 + 23 = 48 ⇒ y = 6

Thus, the required line passes through the point (5, 6).

4x - 2y = 1

2y = 4x - 1

y = 2x – 1/2

Slope of this line = 2

Slope of the required line = -1/2

The required equation of the line is

y - y

y – 6 = -1/2(x - 5)

2y - 12 = -x + 5

x + 2y = 17

Thus, the required line passes through the point (5, 6).

4x - 2y = 1

2y = 4x - 1

y = 2x – 1/2

Slope of this line = 2

Slope of the required line = -1/2

The required equation of the line is

y - y

_{1}= m(x - x_{1})y – 6 = -1/2(x - 5)

2y - 12 = -x + 5

x + 2y = 17