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## Access ML Aggarwal Solutions for Class 10 Maths Chapter 12 Equation of Straight Line

Exercise 12.1

**1. Find the slope of a line whose inclination is
(i) 45°
(ii) 30°**

**Solution: **

The slope of a line having inclination:

(i) 45^{o}

Slope = tan 45^{o} = 1

(ii) 30^{o}

Slope = tan 30^{o} = 1/√3

**2. Find the inclination of a line whose gradient is
(i) 1
(ii) √3**

**(iii) 1/√3**

**Solution: **

Given,

(i) tan θ = 1

⇒ θ = 45^{o}

(ii) tan θ = **√**3

⇒ θ = 60^{o}

(i) tan θ = 1/√3

⇒ θ = 30^{o}

**3. Find the equation of a straight line parallel to x-axis which is at a distance
(i) 2 units above it
(ii) 3 units below it.**

**Solution: **

(i) A line which is parallel to x-axis is y = a

⇒ y = 2

Hence, the equation of line parallel to x-axis which is at a distance of 2 units above it is y – 2 = 0.

(ii) A line which is parallel to x-axis is y = a

⇒ y = -3

Hence, the equation of line parallel to x-axis which is at a distance of 3 units below it is y + 3 = 0.

**4. Find the equation of a straight line parallel to y-axis which is at a distance of:
(i) 3 units to the right
(ii) 2 units to the left.**

**Solution: **

A line which is parallel to y-axis is x = a

(i) Here, x = 3

Hence, the equation of line parallel to y-axis is at a distance of 3 units to the right is x – 3 = 0.

(ii) Here, x = -2

Hence, the equation of line parallel to y-axis at a distance of 2 units to the left is x + 2 = 0.

**5. Find the equation of a straight line parallel to y-axis and passing through the point ( – 3, 5).**

**Solution: **

The equation of the line parallel to y-axis passing through ( – 3, 5) to x = -3

⇒ x + 3 = 0

**6.** **Find the equation of a line whose**

**(i) slope = 3, y-intercept = – 5**

**(ii) slope = -2/7, y-intercept = 3**

**(iii) gradient = √3, y-intercept = -4/3**

**(iv) inclination = 30°, y-intercept = 2**

**Solution: **

Equation of a line whose slope and y-intercept is given by:

y = mx + c, where m is the slope and c is the y-intercept

(i) Given: slope = 3, y-intercept = – 5

⇒ y = 3x + (-5)

Hence, the equation of line is y = 3x – 5.

(ii) Given: slope = -2/7, y-intercept = 3

⇒ y = (-2/7)x + 3

y = (-2x + 21)/7

7y = -2x + 21

Hence, the equation of line is 2x + 7y – 21= 0.

(iii) Given: gradient = √3, y-intercept = -4/3

⇒ y = √3x + (-4/3)

y = (3√3x – 4)/3

3y = 3√3x – 4

Hence, the equation of line is 3√3x – 3y – 4 = 0.

(iv) Given: inclination = 30°, y-intercept = 2

Slope = tan 30^{o} = 1/√3

⇒ y = (1/√3)x + 2

y = (x + 2√3)/ √3

√3y = x + 2√3

Hence, the equation of line is x – √3y + 2√3 = 0.

**7.** **Find the slope and y-intercept of the following lines:**

**(i) x – 2y – 1 = 0 **

**(ii) 4x – 5y – 9 = 0**

**(iii) 3x + 5y + 7 = 0**

**(iv) x/3 + y/4 = 1**

**(v) y – 3 = 0**

**(vi) x – 3 = 0**

**Solution: **

We know that, equation of line whose slope and y-intercept is given by:

y = mx + c, where m is the slope and c is the y-intercept

Using the above and converting to this, we find

(i) x – 2y – 1 = 0

2y = x – 1

⇒ y = (½) x + (-½)

Hence, slope = ½ and y-intercept = – ½

(ii) 4x – 5y – 9 = 0

5y = 4x – 9

⇒ y = (4/5) x + (-9/5)

Hence, slope = 4/5 and y-intercept = -9/5

(iii) 3x + 5y + 7 = 0

5y = -3x – 7

⇒ y = (-3/5) x + (-7/5)

Hence, slope = -3/5 and y-intercept = -7/5

(iv) x/3 + y/4 = 1

(4x + 3y)/ 12 = 1

4x + 3y = 12

3y = -4x + 12

⇒ y = (-4/3) x + 4

Hence, slope = -4/3 and y-intercept = 4

(v) y – 3 = 0

y = 3

⇒ y = (0) x + 3

Hence, slope = 0 and y-intercept = 3

(vi) x – 3 = 0

Here, the slope cannot be defined as the line does not meet y-axis.

**8.** **The equation of the line PQ is 3y – 3x + 7 = 0**

**(i) Write down the slope of the line PQ.**

**(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.**

**Solution: **

Given, equation of line PQ is 3y – 3x + 7 = 0

Re-writing in form of y = mx + c, we have

3y = 3x – 7

⇒ y = x + (-7/3)

Here,

(i) Slope = 1

(ii) As tan θ = 1

θ = 45^{o}

Hence, the angle which PQ makes with the x-axis is Q.

**9. The given figure represents the line y = x + 1 and y = √3x – 1. Write down the angles which the lines make with the positive direction of the x-axis. Hence determine θ.**

**Solution: **

Given line equations, y = x + 1 and y = √3x – 1

On comparing with y = mx + c,

The slope of the line: y = x + 1 is 1 as m = 1

So, tan θ = 1 ⇒ θ = 45^{o}

And,

The slope of the line: y = √3x – 1 is √3 as m = √3

So, tan θ = √3 ⇒ θ = 60^{o}

Now, in triangle formed by the given two lines and x-axis

Ext. angle = Sum of interior opposite angle

60^{o} = θ + 45^{o}

θ = 60^{o} – 45^{o}

Thus, θ = 15^{o}

**10.** **Find the value of p, given that the line y/2 = x – p passes through the point (– 4, 4)**

**Solution: **

Given, equation of line: y/2 = x – p

And, it passes through the point (-4, 4)

Hence, it satisfies the line equation

So,

4/2 = (-4) – p

2 = -4 – p

p = -4 – 2

Thus, p = -6

**11. Given that (a, 2a) lies on the line y/2 = 3x – 6. Find the value of a.**

**Solution: **

Given, equation of line: y/2 = 3x – 6

And, it passes through the point (a, 2a)

Hence, it satisfies the line equation

So,

2a/2 = 3(a) – 6

a = 3a – 6

2a = 6

Thus, a = 3

**12. The graph of the equation y = mx + c passes through the points (1, 4) and (– 2, – 5). Determine the values of m and c.**

**Solution: **

Given, equation of the line is y = mx + c

And, it passes through the points (1, 4)

So, the point will satisfy the line equation

⇒ 4 = m x 1 + c

4 = m + c

m + c = 4 … (i)

Also, the line passes through another point (-2, -5)

So,

5 = m (-2) + c

5 = -2 m + c

⇒ 2m – c = 5 …(ii)

Now, on adding (i) and (ii) we get

3m = 9

⇒ m = 3

Substituting the value of m in (i), we get

3 + c = 4

⇒ c = 4 – 3 = 1

Therefore, m = 3, c = 1

**13. Find the equation of the line passing through the point (2, – 5) and making an intercept of – 3 on the y-axis.**

**Solution: **

Given, a line equation passes through point (2, -5) and makes a y-intercept of -3

We know that,

The equation of line is y = mx + c, where m is the slope and c is the y-intercept

So, we have

y = mx – 3

Now, this line equation will satisfy the point (2, -5)

-5 = m (2) – 3

-5 = 2m – 3

2m = 3 – 5 = -2

⇒ m = -1

Hence, the equation of the line is y = -x + (-3) ⇒ x + y + 3 = 0

**14. Find the equation of a straight line passing through (– 1, 2) and whose slope is 2/5. **

**Solution: **

Given, the equation of straight line passes through (-1, 2) and having slope as 2/5

So, the equation of the line will be

y – y_{1} = m (x – x_{1})

Here, (x_{1}, y_{1}) is (-1, 2)

⇒ y – 2 = (2/5) [x – (-1)]

5 (y – 2) = 2 (x + 1)

5y – 10 = 2x + 2

Thus, the line equation is 2x – 5y + 12 = 0.

**15. Find the equation of a straight line whose inclination is 60° and which passes through the point (0, – 3).**

**Solution:**

Given,

Inclination of a straight line is 60^{o}

So, the slope = tan 60^{o} = √3 = m

And, the equation of line passes through the point (0, -3) = (x_{1}, y_{1})

Hence, the equation of line is given by

y – y_{1} = m (x – x_{1})

y + 3 = √3 (x – 0)

y + 3 = √3x

√3x – y – 3 = 0

Thus, the line equation is √3x – y – 3 = 0.

**16. Find the gradient of a line passing through the following pairs of points.
(i) (0, – 2), (3, 4)
(ii) (3, – 7), (– 1, 8)**

**Solution: **

Gradient of a line (m) = y_{2 }– y_{1} / x_{2 }– x_{1}

(i) (0, – 2), (3, 4)

m = (4 + 2)/(3 – 0) = 6/3 = 2

Hence, gradient = 2

(ii) (3, – 7), (– 1, 8)

m = (8 + 7)/(-1 – 3) = 15/-4

Hence, gradient = -15/4

**17. The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find:
(i) The gradient of EF
(ii) The equation of EF
(iii) The coordinates of the point where the line EF intersects the x-axis.**

**Solution: **

Given, co-ordinates of points E and F are (0, 4) and (3, 7) respectively

(i) The gradient of EF

m = y_{2 }– y_{1} / x_{2 }– x_{1} = (7 – 4)/(3 – 0) = 3/3

⇒ m = 1

(ii) Equation of line EF is given by,

y – y_{1} = m (x – x_{1})

y – 7 = 1 (x – 3)

y – 7 = x – 3

x – y + 7 – 3 = 0

Hence, the equation of line EF is x – y + 4 = 0

(iii) It’s seen that the co-ordinates of point of intersection of EF and the x-axis will be y = 0

So, substituting the value y = 0 in the above equation

x – y + 4 = 0

x – 0 + 4 = 0

x = -4

Hence, the co-ordinates are (-4, 0).

**18. Find the intercepts made by the line 2x – 3y + 12 = 0 on the co-ordinate axis.**

**Solution: **

Given line equation is 2x – 3y + 12 = 0

On putting y = 0, we will get the intercept made on x-axis

2x – 3y + 12 = 0

2x – 3 × 0 + 12 = 0,

2x – 0 + 2 = 0,

2x = -12

⇒ x = -6

Now, on putting x = 0, we get the intercepts made on y-axis

2x – 3y + 12 = 0

2 × 0 – 3y + 12 = 0

-3y = -12

⇒ y = 4

Hence, the x-intercept and y-intercept of the given line is -6 and 4 respectively.

**19. Find the equation of the line passing through the points P (5, 1) and Q (1, – 1). Hence, show that the points P, Q and R (11, 4) are collinear.**

**Solution: **

Given, two points P (5, 1) and G (1, -1)

Slope of the line (m) = y_{2} – y_{1}/ x_{2} – x_{1}

= -1 – 1/ 1 – 5

= -2/-4 = ½

So, the equation of the line is

y – y_{1} = m (x – x_{1})

y – 1 = ½ (x – 5)

2y – 2 = x – 5

x – 2y – 3 = 0

Now, if point R (11, 4) is collinear to points P and Q then, R (11, 4) should satisfy the line equation

On substituting, we have

11 – 2(4) – 3 = 11 – 8 – 3 = 0

As point R satisfies the line equation,

Hence, P, Q and R are collinear.

**20. 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.**

**Solution: **

Given,

Points A (a, 3), B (2,1) and C (5, a) are collinear.

So, slope of AB = slope of BC

⇒ -6 = (a – 1) (2 – a) [On cross multiplying]

-6 = 2a – 2 – a^{2} + a

-6 = 3a – a^{2} – 2

a^{2} – 3a – 4 = 0

a^{2} – 4a + a – 4 = 0

a (a – 4) + (a – 4) = 0

(a + 1) (a – 4) = 0

a = -1 or 4

As a = -1 doesn’t satisfy the equation

⇒ a = 4

Now,

Slope of BC = (a – 1)/(5 – 2) = (4 – 1)/3 = 3/3 = 1 = m

So, the equation of BC is

(y – 1) = 1 (x – 2)

y – 1 = x – 2

x – y = -1 + 2

Thus, the equation of BC is x – y = 1.

**21. Use a graph paper for this question. The graph of a linear equation in x and y, passes through A (– 1, – 1) and B (2, 5). From your graph, find the values of h and k, if the line passes through (h, 4) and (½, k).**

**Solution: **

Given,

Points (h, 4) and (1/2, k) lie on the line passing through A (-1, -1) and B (2, 5)

From the graph, its clearly seen that

h = 3/2 and

k = 2

**22. ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, – 4). Find
(i) the coordinates of A
(ii) the equation of the diagonal BD.**

**Solution: **

(i) Given,

ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4)

O in the point of intersection of the diagonals of the parallelogram.

So, the co-ordinates of O = ([5 + 2]/2, [8 – 4]/2) = (3.5, 2)

Now, for the line AC we have

3.5 = (x + 4)/2 and 2 = (y + 7)/2

7 = x + 4 and 4 = y + 7

x = 7 – 4 and y = 4 – 7

x = 3 and y = -3

Thus, the co-ordinates of A are (3, -3).

(ii) Equation of diagonal BD is given by

y – 8 = (-4 – 8)/(2 – 5) × (x – 5)

y – 8 = (-12/-3) × (x – 5)

y – 8 = 4 (x – 5)

y – 8 = 4x – 20

4x – y – 20 + 8 = 0

Hence, the equation of the diagonal is 4x – y – 12 = 0.

**23. In ∆ABC, A (3, 5), B (7, 8) and C (1, – 10). Find the equation of the median through A.**

**Solution: **

Given,

∆ABC and their vertices A (3, 5), B (7, 8) and C (1, – 10).

And, AD is median

So, D is mid-point of BC

Hence, the co-ordinates of D is ([7 + 1]/2, [8 – 10]/2) = (4, -1)

Now,

Slope of AD, m = y_{2} – y_{1}/ x_{2} – x_{1}

m = (5 + 1)/ (3 – 4) = 6/-1 = -6

Thus, the equation of AD is given by

y – y_{1} = m (x – x_{1})

y + 1 = -6 (x – 4)

y + 1 = -6x + 24

⇒ 6x + y – 23 = 0

**24. Find the equation of a line passing through the point (– 2, 3) and having x-intercept 4 units.**

**Solution: **

Given, point (-2, 3) and the x-intercept of the line passing through that point is 4 units.

So, the co-ordinates of the point where the line meets the x-axis is (4, 0)

Now, slope of the line passing through the points (-2, 3) and (4, 0)

m = y_{2} – y_{1}/ x_{2} – x_{1}

m = (0 – 3)/(4 + 2) = -3/6 = -1/2

Hence, the equation of the line will be

y – y_{1} = m (x – x_{1})

y – 0 = -½ (x – 4)

2y = -x + 4

⇒ x + 2y = 4

**25. Find the equation of the line whose x-intercept is 6 and y-intercept is – 4.**

**Solution: **

Given, x-intercept of a line is 6

So,

The line will pass through the point (6, 0)

Also given, the y -intercept of the line is -4 ⇒ c = -4

So, the line will pass through the point (0, -4)

Now,

Slope, m = m = y_{2} – y_{1}/ x_{2} – x_{1}

m = (-4 – 0)/(0 – 6) = -4/-6 = 2/3

Thus, the equation of the line is given by

y = mx + c

y = (2/3)x + (-4)

3y = 2x – 12

⇒ 2x – 3y – 12 = 0

**26. Write down the equation of the line whose gradient is 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.**

**Solution: **

Given, P divides the line segment joining the points A (-2, 6) and B (3, -4) in the ratio 2: 3

So, the co-ordinates of P will be

x = (m_{1}x_{2} + m_{2}x_{1})/(m_{1} + m_{2})

= (2×3 + 3×(-2))/ (2 + 3)

= (6 – 6)/5

= 0/5 = 0

y = (m_{1}y_{2} + m_{2}y_{1})/(m_{1} + m_{2})

= (2×(-4) + 3×(6))/ (2 + 3)

= (-8 + 18)/5 = 10/5

= 2

Hence, the co-ordinates of P are (0, 2)

Now, the slope (m) of the line passing through (0, 2) is 3/2

Thus, the equation will be

y – y_{1} = m (x – x_{1})

y – 2 = 3/2 (x – 0)

2y – 4 = 3x

⇒ 3x – 2y + 4 = 0

**27. Find the equation of the line passing through the point (1, 4) and intersecting the line x – 2y – 11 = 0 on the y-axis.**

**Solution: **

Given, line x – 2y – 11 = 0 passes through y-axis and point (1, 4)

So, putting x = 0 in the line equation we get the y-intercept

0 – 2y – 11 = 0

y = -11/2

The co-ordinates are (0, -11/2)

Now, the slope of the line joining the points (1, 4) and (0, -11/2) is given by

m = y_{2} – y_{1}/ x_{2} – x_{1}

= (-11/2 – 4)/ (0 – 1)

= 19/2

Thus, the line equation will be

y – y_{1} = m (x – x_{1})

y + 11/2 = 19/2 (x – 0)

2y + 11 = 19x

⇒ 19x – 2y – 11 = 0

**28. Find the equation of the straight line containing the point (3, 2) and making positive equal intercepts on axes.**

**Solution: **

Let the line containing the point P (3, 2) pass through x-axis at A (x, 0) and y-axis at B (0, y)

Given, OA = OB

Thus, x = y

Now, the slope of the line (m) = y_{2} – y_{1}/ x_{2} – x_{1}

= 0 – y/ x – 0

= -x/x = -1

Hence, the equation of the line will be

y – y_{1} = m (x – x_{1})

y – 2 = -1 (x – 3)

y – 2 = -x + 3

⇒ x + y – 5 = 0

**29. Three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2) find:
(i) the coordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD.**

**Solution: **

Given, the three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2)

(i) We know that the diagonals of a parallelogram bisect each other.

Let (x, y) be the co-ordinates of D

Hence, we have

Mid-point of diagonal AC = ((3 + 3)/2, (6 + 2)/2) = (3, 4)

Mid-point of diagonal BD = ((5 + x)/2, (10 + y)/2)

And, these two should be the same

On equating we get,

(5 + x)/2 = 3 and (10 + y)/2 = 4

5 + x = 6 and 10 + y = 8

x = 1 and y = -2

Thus, the co-ordinates of D = (1, -2)

(ii) Length of diagonal BD

(iii) Equation of the side joining A (3, 6) and D (1, -2) is given by

4 (x – 3) = y – 6

4x – 12 = y – 6

4x – y = 6

Thus, the equation of the side joining A (3, 6) and D (1, -2) is 4x – y = 6.

**30. 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) the co-ordinates of A and B.
(ii) the slope of the line AB.
(iii) the equation of the line AB.**

**Solution: **

Given, points A and B are on x-axis and y-axis respectively

Let co-ordinates of A be (x, 0) and of B be (0, y)

And P (2, -3) is the midpoint of AB

So, we have

2 = (x + 0)/2 and -3 = (0 + y)/2

x = 4 and y = -6

(i) Hence, the co-ordinates of A are (4, 0) and of B are (0, -6).

(ii) Slope of AB = y_{2} – y_{1}/ x_{2} – x_{1}

= (-6 – 0)/ (0 – 4)

= -6/-4 = 3/2 = m

(iii) Equation of AB will be

y – y_{1} = m (x – x_{1})

y – (-3) = 3/2 (x – 2) [As P lies on it]

y + 3 = 3/2 (x – 2)

2y + 6 = 3x – 6

3x – 2y – 12 = 0

**31. Find the equations of the diagonals of a rectangle whose sides are x = – 1, x = 2, y = – 2 and y = 6.**

**Solution: **

Given,

The equations of sides of a rectangle are

x_{1} = -1, x_{2} = 2, y_{1} = -2, y_{2} = 6.

These lines form a rectangle when they intersect at A, B, C, D respectively

Now,

The co-ordinates of A, B, C and D will be (-1, -2), (2, -2), (2, 6) and (-1, 6) respectively.

And, AC and BD are its diagonals

Slope of the diagonal AC

= y_{2} – y_{1}/ x_{2} – x_{1}

= (6 + 2)/ (2 + 1)

= 8/3 = m

So, the equation of AC will be

y – y_{1} = m (x – x_{1})

y + 2 = 8/3 (x + 1)

3y + 6 = 8x + 8

⇒ 8x – 3y + 2 = 0

**32. Find the equation of a straight line passing through the origin and through the point of intersection of the lines 5x + 1y – 3 and 2x – 3y = 7**

**Solution: **

Given line equations,

5x + 7y = 3 … (i)

2x – 3 y = 7 … (ii)

Now, performing multiplication of (i) by 3 and (ii) by 7, we get

15x + 21y = 9

14x – 21y = 49

On adding we get,

29x = 58

x = 58/29 = 2

Substituting the value of x in (i), we get

5(2) + 7y = 3

10 + 7y = 3

7y = 3 – 10

y = -7/7 = -1

Hence, the point of intersection of lines is (2, -1)

Now, the slope of the line joining the points (2, -1) and (0, 0) will be

m = y_{2} – y_{1}/ x_{2} – x_{1}

= (0 + 1)/ (0 – 2)

= -1/2

Equation of the line is given by:

y – y_{1} = m (x – x_{1})

y – 0 = -1/2 (x – 0)

2y = -x

Thus, the required line equation is x + 2y = 0.

**33. Point A (3, – 2) on reflection in the x-axis is mapped as A’ and point B on reflection in the y-axis is mapped onto B’ ( – 4, 3).
(i) Write down the co-ordinates of A’ and B.
(ii) Find the slope of the line A’B, hence find its inclination.**

**Solution: **

Given,

A’ is the image of A (3, -2) on reflection in the x-axis.

(i) The co-ordinates of A’ will be (3, 2).

Again B’ (- 4, 3) in the image of A’, when reflected in the y-axis

Hence, the co-ordinates of B will be (4, 3)

(ii) Slope of the line joining, the points A’ (3, 2) and B (4, 3) will be

m = y_{2} – y_{1}/ x_{2} – x_{1}

= (2 – 3)/ (3 – 4)

= -1/-1 = 1

So, tan θ = 45^{o}

Thus, the angle of inclination is 45^{o}.

Exercise 12.2

**1.** **State which one of the following is true: The straight lines y = 3x – 5 and 2y = 4x + 7 are**

**(i) parallel**

**(ii) perpendicular**

**(iii) neither parallel nor perpendicular.**

**Solution: **

Given straight lines: y = 3x – 5 and 2y = 4x + 7 ⇒ y = 2x + 7/2

And, their slopes are 3 and 2

The product of slopes is 3 x 2 = 6.

Hence, as the slopes of both the lines are neither equal nor their product is -1 the given pair of straight lines are neither parallel nor perpendicular.

**2. If 6x + 5y – 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.**

**Solution: **

For two lines to be parallel, their slopes must be same.

Given line equations,

6x + 5y – 7 = 0 and 2px + 5y + 1 = 0

In equation 6x + 5y – 7 = 0,

5y = -6x + 7

y = (-6/5) x + 7/5

So, the slope of the line (m_{1}) = -6/5

Again, in equation 2px + 5y + 1 = 0

5y = -2px – 1

y = (-2p/5) x – 1/5

So, the slope of the line (m_{2}) = -2p/5

For these two lines to be parallel

m_{1} = m_{2}

-6/5 = -2p/5

p = (-6/5) x (-5/2)

Thus, p = 3

**3. Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel. Find the relation connecting a and b.**

**Solution: **

Given lines are: 2x – by + 5 = 0 and ax + 3y = 2

If two lines to be parallel then their slopes must be equal.

In equation 2x – by + 5 = 0,

by = 2x + 5

y = (2/b) x + 5/b

So, the slope of the line (m_{1}) = 2/b

And in equation ax + 3y = 2,

3y = -ax + 2

y = (-a/3) x + 2/3

So, the slope of the line (m_{2}) = (-a/3)

As the lines are parallel

m_{1} = m_{2}

2/b = -a/3

6 = -ab

Hence, the relation connecting a and b is ab + 6 = 0

**4. Given that the line y/2 = x – p and the line ax + 5 = 3y are parallel, find the value of a.**

**Solution: **

Given,

Line equation: y/2 = x – p

⇒ y = 2x – 2p

Here, the slope of the line is 2.

And, another line equation: ax + 5 = 3y

⇒ 3y = ax + 5

y = (a/3) x + 5/3

Hence, the slope of the line is a/3

As the line are parallel, their slopes must be equal

⇒ 2 = a/3

a = 6

Thus, the value of a is 6.

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

**Solution: **

If two lines are perpendicular, then the product of their slopes is -1

Now, slope of the line y = 3x + 7 is m_{1} = 3

And,

The slope of the line: 2y + px = 3

2y = -px + 3

y = (-p/2) x + 3

m_{2} = -p/2

As the lines as perpendicular,

⇒ m_{1} x m_{2} = -1

3 x (-p/2) = -1

p = 2/3

Thus, the value of p is 2/3.

**6. If the straight lines kx – 5y + 4 = 0 and 4x – 2y + 5 = 0 are perpendicular to each other. Find the value of k. **

**Solution: **

Given,

In equation, kx – 5y + 4 = 0

⇒ 5y = kx + 4

y = (k/5) x + 4/5

So, the slope (m_{1}) = k/5

And, in equation 4x – 2y + 5 = 0

⇒ 2y = 4x + 5

y = 2x + 5/2

So, the slope (m_{2}) = 2

As the lines are perpendicular to each other

⇒ m_{1} x m_{2} = -1

k/5 x 2 = -1

k = (-1 x 5)/2

Hence, the value of k = -5/2

**7. If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.**

**Solution: **

Given that the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other

Then the product of their slopes must be -1.

Slope of line 3x + by + 5 = 0 is,

by = -3x – 5

y = (-3/b) – 5/b

So, slope (m_{1}) = -3/b

And,

The slope of line ax – 5y + 7 = 0 is

5y = ax + 7

y = (a/5) x + 7/5

So, slope (m_{2}) = a/5

As the lines are perpendicular, we have

m_{1} x m_{2} = -1

-3/b x a/5 = -1

-3a/5b = -1

-3a = – 5b

3a = 5b

Hence, the relation connecting a and b is 3a = 5b.

**8. Is the line through (– 2, 3) and (4, 1) perpendicular to the line 3x = y + 1?
Does the line 3x = y + 1 bisect the join of (– 2, 3) and (4, 1). **

**Solution: **

Slope of the line passing through the points (-2, 3) and (4, 1) is given by

m_{1 }= y_{2} – y_{1}/ x_{2} – x_{1}

= (1 – 3)/ (4 + 2)

= -2/6

= -1/3

And, the slope of the line: 3x = y + 1

y = 3x -1

Slope (m_{2}) = 3

Now,

m_{1} x m_{2 }= -1/3 x 3 = -1

Thus, the lines are perpendicular to each other as the product of their slopes is -1.

Now,

Co-ordinates of the mid-point of the line joining the points (-2, 3) and (4, 1) is

([-2 + 4]/2, [3 + 1]/2) = (1, 2)

Now, if the line 3x = y + 1 passes through the mid-point then it will satisfy the equation

3(1) = (2) + 1

3 = 3

Hence, the line 3x = y + 1 bisects the line joining the points (– 2, 3) and (4, 1).

**9. The line through A (– 2, 3) and B (4, b) is perpendicular to the line 2x – 4y = 5. Find the value of b.**

**Solution: **

**The slope of the line passing through A (-2, 3) and B (4, b) will be **

m_{1} = (b – 3)/ (4 + 2) = (b – 3)/ 6

Now, the gradient of the given line 2x – 4y = 5 is

4y = 2x + 5

y = (2/4) x + 5/4

y = ½ x + 5/4

So, m_{2} = ½

As the line are perpendicular to each other, we have

m_{1} x m_{2} = -1

(b – 3)/ 6 × ½ = -1

(b – 3)/ 12 = -1

b – 3 = -12

b = -12 + 3 = -9

Hence, the value of b is -9.

**10. If the lines 3x + y = 4, x – ay + 7 = 0 and bx + 2y + 5 = 0 form three consecutive sides of a rectangle, find the value of a and b.**

**Solution: **

Given lines are:

3x + y = 4 … (i)

x – ay + 7 = 0 … (ii)

bx + 2y + 5 = 0 … (iii)

It’s said that these lines form three consecutive sides of a rectangle.

So,

Lines (i) and (ii) must be perpendicular

Also, lines (ii) and (iii) must be perpendicular

We know that, for two perpendicular lines the product of their slopes will be -1.

Now,

Slope of line (i) is

3x + y = 4 ⇒ y = -3x = 4

Hence, slope (m_{1}) = -3

And, slope of line (ii) is

x – ay + 7 = 0 ⇒ ay = x + 7

y = (1/a) x + 7/a

Hence, slope (m_{2}) = 1/a

Finally, the slope of line (iii) is

bx + 2y + 5 = 0 ⇒ 2y = -bx – 5

y = (-b/2) x – 5/2

Hence, slope (m_{3}) = -b/2

As lines (i), (ii) and (iii) are consecutive sides of rectangle, we have

m_{1} x m_{2} = -1 and m_{2} x m_{3} = -1

(-3) x (1/a) = -1 and (1/a) x (-b/2) = -1

-3 = -a and -b/2a = -1

a = 3 and b = 2a ⇒ b = 2(3) = 6

Thus, the value of a is 3 and the value of b is 6.

**11. Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x – 3y – 7 = 0. Find the coordinates of the point where it cuts the x-axis.**

**Solution: **

Given line: 2x – 3y – 7 = 0

Its slope is,

3y = 2x – 7

y = (2/3) x – 7/3

⇒ m = 2/3

So, the equation of the line parallel to the given line will be 2/3

Also given, the y-intercept is 4 = c

Hence, the equation of the line is given by

y = mx + c

y = (2/3) x + 4

3y = 2x + 12

2x – 3y + 12 = 0

Now, when this line intersects the x-axis the y co-ordinate becomes zero.

So, putting y = 0 in the line equation, we get

2x – 3(0) + 12 = 0

2x + 12 = 0

x = -12/2 = 6

Hence, the co-ordinates of the point where it cuts the x-axis is (-6, 0).

**12. Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with y-intercept – 3 units.**

**Solution: **

Given line: 2x + 5y + 7 = 0

So, its slope is given by

5y = -2x – 7

y = (-2/5) – 7/5

⇒ m = -2/5

Now, let the slope of the line perpendicular to this line be m’

Then,

m x m’ = -1

(-2/5) x m’ = -1

⇒ m’ = 5/2

Also given, the y-intercept (c) = -3

Hence, the equation of the line is given by

y = m’x + c

y = (5/2) x + (-3)

2y = 5x – 6

5x – 2y – 6 = 0

**13. Find the equation of a st. line perpendicular to the line 3x – 4y + 12 = 0 and having same y-intercept as 2x – y + 5 = 0.**

**Solution: **

Given line: 3x – 4y + 12 = 0

The slope of the line is given by

3x – 4y + 12 = 0 ⇒ 4y = 3x + 12

y = (3/4) x + 3

Thus, slope (m_{1}) = ¾

Now, let the slope of the line perpendicular to the given line be taken as m_{2}

So,

m_{1} x m_{2} = -1

(3/4) x m_{2} = -1

m_{2} = -4/3

And given, the y-intercept of the line is same as 2x – y + 5 = 0

⇒ y = 2x + 5

So, the y-intercept is 5 = c.

Hence, the equation of line is given by

y = m_{2}x + c

y = (-4/3) x + 5

3y = -4x + 15

4x + 3y = 15

**14. Find the equation of the line which is parallel to 3x – 2y = – 4 and passes through the point (0, 3).**

**Solution: **

Given line: 3x – 2y = -4

Slope (m_{1}) is given by

2y = 3x + 4

y = (3/2) x + 2

So, m_{1} = 3/2

Now, the slope of the line parallel to the given line will have the same slope as 3/2 = m

And the line passes through point (0, 3)

Thus, the equation of the required line is given by

y = mx + c

y = (3/2) x + 3

2y = 3x + 6

3x – 2y + 6 = 0

**15. Find the equation of the line passing through (0, 4) and parallel to the line 3x + 5y + 15 = 0.**

**Solution: **

Given line: 3x + 5y + 15 = 0

5y = -3x – 15

y = (-3/5) x – 3

So, slope (m) = -3/5

The slope of the line parallel to the given line will the same -3/5

And, the line passes through the point (0, 4)

Hence, equation of the line will be

y – y_{1} = m (x – x_{1})

y – 4 = (-3/5) (x – 0)

5y – 20 = -3x

3x + 5y – 20 = 0

**16. The equation of a line is y = 3x – 5. Write down the slope of this line and the intercept made by it on the y-axis. Hence or otherwise, write down the equation of a line which is parallel to the line and which passes through the point (0, 5).**

**Solution: **

Given line: y = 3x – 5

Here slope (m_{1}) = 3

Substituting x = 0, we get y = – 5

Hence, the y-intercept = – 5

Now, the slope of the line parallel to the given line will be 3 and it passes through the point (0, 5).

Thus, equation of the line will be

y – y_{1} = m (x – x_{1})

y – 5 = 3 (x – 0)

y = 3x + 5

**17. Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point (– 1, – 2).**

**Solution: **

Given line: 3x + 8y = 12

8y = -3x + 12

y = (-3/8) x + 12

So, the slope (m_{1}) = -3/8

Let’s consider the slope of the line perpendicular to the given line as m_{2}

Then, m_{1} x m_{2} = -1

-3/8 x m_{2} = -1

m_{2} = 8/3

Now,

The equation of the line perpendicular to the given line and passing through the point (-1, -2) will be

y – y_{1} = m (x – x_{1})

y – (-2) = (8/3) (x – (-1))

y + 2 = (8/3) (x + 1)

3y + 6 = 8x + 8

3y = 8x + 2

Thus, the equation of the required line is 3y = 8x + 2.

**18. (i) The line 4x – 3y + 12 = 0 meets the x-axis at A. Write down the co-ordinates of A.
(ii) Determine the equation of the line passing through A and perpendicular to 4x – 3y + 12 = 0.**

**Solution: **

Given line: 4x – 3y + 12 = 0

(i) When this line meets the x-axis, its y co-ordinate becomes 0.

So, putting y = 0 in the given equation, we get

4x – 3(0) + 12 = 0

4x + 12 = 0

x = -12/4

x = -3

Hence, the line meets the x-axis at A (-3, 0).

(ii) Now, the slope of the line is given by

4x – 3y + 12 = 0

3y = 4x + 12

y = (4/3) x + 4

⇒ m_{1} = 4/3

Let’s assume the slope of the line perpendicular to the given line be m_{2}

Then, m_{1} x m_{2} = -1

4/3 x m_{2} = -1

m_{2} = -3/4

Thus, the equation of the line perpendicular to the given line passing through A will be

y – 0 = -3/4 (x + 3)

4y = -3(x + 3)

3x + 4y + 9 = 0

**19. Find the equation of the line that is parallel to 2x + 5y – 7 = 0 and passes through the mid-point of the line segment joining the points (2, 7) and (– 4, 1).**

**Solution:**

Given line: 2x + 5y – 7 = 0

5y = -2x + 7

y = (-2/5) x + 7/5

So, the slope is -2/5

Hence, the slope of the line that is parallel to the given line will be the same, m = -2/5

Now, the mid-point of the line segment joining points (2, 7) and (– 4, 1) is

((2 – 4)/2, (7 + 1)/2) = (-1, 4)

Thus, the equation of the line will be

y – y_{1} = m (x – x_{1})

y – 4 = (-2/5) (x + 1)

5y – 20 = -2x -2

2x + 5y = 18

**20. Find the equation of the line that is perpendicular to 3x + 2y – 8 = 0 and passes through the mid-point of the line segment joining the points (5, – 2) and (2, 2).**

**Solution: **

Given line: 3x + 2y – 8 = 0

2y = -3x + 8

y = (-3/2) x + 4

Here, slope (m_{1}) = -3/2

Now, the co-ordinates of the mid-point of the line segment joining the points (5, -2) and (2, 2) will be

((5 + 2)/7, (-2 + 2)/7) = (7/2, 0)

Let’s consider the slope of the line perpendicular to the given line be m_{2}

Then,

m_{1} x m_{2} = -1

(-3/2) x m_{2} = -1

m_{2} = 2/3

So, the equation of the line with slope m_{2} and passing through (7/2, 0) will be

y – 0 = (2/3) (x – 7/2)

3y = 2x – 7

2x – 3y – 7 = 0

Thus, the required line equation is 2x – 3y – 7 = 0.

**21. Find the equation of a straight line passing through the intersection of 2x + 5y – 4 = 0 with x-axis and parallel to the line 3x – 7y + 8 = 0.**

**Solution: **

Let’s assume the point of intersection of the line 2x + 5y – 4 = 0 and x-axis be (x, 0)

Now, substituting the value y = 0 in the line equation, we have

2x + 5(0) – 4 = 0

2x – 4 = 0

x = 4/2 = 2

Hence, the co-ordinates of the point of intersection is (2, 0)

Also given, line equation: 3x – 7y + 8 = 0

7y = 3x + 8

y = (3/7) x + 8/7

So, the slope (m) = 3/7

We know that the slope of any line parallel to the given line will be the same.

So, the equation of the line having slope 3/7 and passing through the point (2, 0) will be

y – 0 = (3/7) (x – 2)

7y = 3x – 6

3x – 7y – 6 = 0

Thus, the required line equation is 3x – 7y – 6 = 0.

**22. 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 equation: 3x + 4y – 7 = 0

(i) Slope of the line is given by,

4y = -3x + 7

y = (-3/4) x + 7

Hence, slope (m_{1}) = -3/4

(ii) Let the slope of the perpendicular to the given line be m_{2}

Then, m_{1} x m_{2} = -1

(-3/4) x m_{2} = -1

m_{2} = 4/3

Now, to find the point of intersection of

x – y + 2 = 0 … (i)

3x + y – 10 = 0 … (ii)

On adding (i) and (ii), we get

4x – 8 = 0

4x = 8

x = 8/4 = 2

Putting x = 2 in (i), we get

2 – y + 2 = 0

y = 4

Hence, the point of intersection of the lines is (2, 4)

The equation of the line having slope m_{2} and passing through (2, 4) will be

y – 4 = (4/3) (x – 2)

3y – 12 = 4x – 8

4x – 3y + 4 = 0

Thus, the required line equation is 4x – 3y + 4 = 0.

**23. Find the equation of the line perpendicular from the point (1, – 2) on the line 4x – 3y – 5 = 0. Also find the co-ordinates of the foot of perpendicular.**

**Solution: **

Given line equation: 4x – 3y – 5 = 0

3y = 4x – 5

y = (4/3) x – 5

Slope of the line (m_{1}) = 4/3

Let the slope of the line perpendicular to the given line be m_{2}

Then, m_{1} x m_{2} = -1

(4/3) x m_{2} = -1

m_{2} = -3/4

Now, the equation of the line having slope m_{2} and passing through the point (1, -2) will be

y + 2 = (-3/4) (x – 1)

4y + 8 = -3x + 3

3x + 4y + 5 = 0

Next, for finding the co-ordinates of the foot of the perpendicular which is the point of intersection of the lines

4x – 3y – 5 = 0 …. (1) and

3x + 4y + 5 = 0 …. (2)

On multiplying (1) by 4 and (2) by 3, we get

16x – 12y – 20 = 0

9x + 12y + 15 = 0

Adding we get,

25x – 5 = 0

x = 5/25

x = 1/5

Putting the value of x in (1), we have

4(1/5) – 3y – 5 = 0

4/5 – 3y – 5 = 0

3y = 4/5 – 5 = (4 – 25)/5

3y = -21/5

y = -7/5

Thus, the co-ordinates are (1/5, -7/5)

**24. Prove that the line through (0, 0) and (2, 3) is parallel to the line through (2, – 2) and (6, 4).**

**Solution: **

Let the slope of the line through (0, 0) and (2, 3) be m_{1}

So, m_{1} = (y_{2} – y_{1})/ (x_{2} – x_{1})

= (3 – 0)/ (2 – 0)

= 3/2

And, let the slope of the line through (2, -2) and (6, 4) be m_{2}

So, m_{2} = (y_{2} – y_{1})/ (x_{2} – x_{1})

= (4 + 2)/ (6 – 2)

= 6/4 = 3/2

It’s clearly seen that the slopes m_{1} = m_{2}

Thus, the lines are parallel to each other.

**25. Prove that the line through (– 2, 6) and (4, 8) is perpendicular to the line through (8, 12) and (4, 24).**

**Solution: **

Let the slope of the line through points (– 2, 6) and (4, 8) be m_{1}

So, m_{1} = (y_{2} – y_{1})/ (x_{2} – x_{1})

= (8 – 6)/ (4 + 2)

= 2/6

= 1/3

And, let the slope of the line through (8, 12) and (4, 24) be m_{2}

So, m_{2} = (y_{2} – y_{1})/ (x_{2} – x_{1})

= (24 – 12)/ (4 – 8)

= 12/ (-4)

= -3

Now, product of slopes is

m_{1} x m_{2} = 1/3 x (-3) = -1

Thus, the lines are perpendicular to each other.

**26. Show that the triangle formed by the points A (1, 3), B (3, – 1) and C (– 5, – 5) is a right-angled triangle by using slopes.**

**Solution: **

Given, points A (1, 3), B (3, – 1) and C (– 5, – 5) form a triangle

Now,

Slope of the line AB = m_{1} = (-1 – 3)/ (3 – 1) = -4/2 = -2

And,

Slope of the line BC = m_{2} = (-5 + 1)/ (-5 – 3) = -4/-8 = ½

Hence,

m_{1} x m_{2} = (-2) x (1/2) = -1

So, the lines AB and BC are perpendicular to each other.

Therefore, ∆ABC is a right-angled triangle.

**27. Find the equation of the line through the point (– 1, 3) and parallel to the line joining the points (0, – 2) and (4, 5).**

**Solution: **

Slope of the line joining the points (0, -2) and (4, 5) is

m = (5 + 2)/ (4 – 0)

= 7/4

Now, the slope of the line parallel to it and passing through (-1, 3) will be also be 7/4

Hence, the equation of the line is

y – y_{1} = m (x – x_{1}) ⇒ y – 3 = 7/4 (x + 1)

4y – 12 = 7x + 7

7x – 4y + 19 = 0

**28. are the vertices of a triangle. **

**(i) Find the coordinates of the centroid G of the triangle. **

**(ii) Find the equation of the line through G and parallel to AC. **

**Solution: **

Given, A (– 1, 3), B (4, 2), C (3, – 2)

(i) Co-ordinates of centroid G is

G (x, y) = ((x_{1} + x_{2} + x_{3})/2, (y_{1}+ y_{2} + y_{3})/2)

= ((-1 + 4 + 3)/3, (3 + 2 – 2)/3)

= (6/3, 3/3) = (2, 1)

Hence, the co-ordinates of the centroid G of the triangle is (2, 1)

(ii) Slope of AC = (y_{2} – y_{1})/ (x_{2} – x_{1}) = (-2 – 3)/ (3 – (-1)) = -5/4

So, the slope of the line parallel to AC is also -5/4

Now, the equation of line through G is

y – 1 = (-5/4) (x – 2)

4y – 4 = -5x + 10

5x + 4y = 14

Thus, the required line equation is 5x + 4y = 14.

**29. The line through P (5, 3) intersects y-axis at Q. (i) Write the slope of the line. (ii) Write the equation of the line. (iii) Find the coordinates of Q.**

**Solution: **

(i) Here, θ = 45^{o}

So, the slope of the line = tan θ = tan 45^{o} = 1

(ii) Equation of the line through P and Q is

y – 3 = 1(x – 5)

x – y – 2 = 0

(iii) Let the co-ordinates of Q be (0, y)

Then, m = y_{2} – y_{1}/ x_{2} – x_{1}

1 = (3 – y)/ (5 – 0)

5 = 3 – y

y = 3 – 5 = -2

Thus, co-ordinates of Q are (0, -2).

**30. In the adjoining diagram, write down (i) the co-ordinates of the points A, B and C. (ii) the equation of the line through A parallel to BC.**

**Solution: **

From the given figure, its clearly seen that

Co-ordinates of A are (2, 3) and of B are (-1, 2) and of C are (3, 0).

Now,

Slope of BC = (0 – 2)/ (3 – (-1))

= -2/4

= -1/2

So, the slope of the line parallel to BC is also -1/2

And, the line passes through A (2, 3)

Hence, the equation will be

y – 3 = (-1/2) (x – 2)

2y – 6 = -x + 2

x + 2y = 8

**31. Find the equation of the line through (0, – 3) and perpendicular to the line joining the points (– 3, 2) and (9, 1).**

**Solution: **

The slope of the line joining the points (-3, 2) and (9, 1) is

m_{1} = (1 – 2)/ (9 + 3) = -1/12

Now, let the slope of the line perpendicular to the above line be m_{2}

Then, m_{1} x m_{2} = -1

(-1/12) x m_{2} = -1

m_{2} = 12

So, the equation of the line passing through (0, -3) and having slope of m_{2} will be

y – (-3) = 12 (x – 0)

y + 3 = 12x

12x – y = 3

Thus, the required line equation is 12x – y = 3.

**32. The vertices of a triangle are A (10, 4), B (4, – 9) and C (– 2, – 1). Find the equation of the altitude through A. The perpendicular drawn from a vertex of a triangle to the opposite side is called altitude.**

**Solution: **

Given, vertices of a triangle are A (10, 4), B (4, – 9) and C (– 2, – 1)

Now,

Slope of line BC (m_{1}) = (-1 + 9)/ (-2 – 4) = 8/ (-6) = -4/3

Let the slope of the altitude from A (10, 4) to BC be m_{2}

Then, m_{1} x m_{2} = -1

(-4/3) x m_{2} = -1

m_{2} = ¾

So, the equation of the line will be

y – 4 = ¾ (x – 10)

4y – 16 = 3x – 30

3x – 4y – 14 = 0

**33. A (2, – 4), B (3, 3) and C (– 1, 5) are the vertices of triangle ABC. Find the equation of: **

**(i) the median of the triangle through A **

**(ii) the altitude of the triangle through B.**

**Solution: **

Given, A (2, – 4), B (3, 3) and C (– 1, 5) are the vertices of triangle ABC

(i) D is the mid-point of BC

So, the co-ordinates of D will be

((3 – 1)/2, (3 + 5)/2) = (2/2, 8/2) = (1, 4)

Now,

The slope of AC (m_{1}) = (5 + 4)/ (-1 – 2) = 9/-3 = -3

Let the slope of BE be m_{2}

Then, m_{1} x m_{2} = -1

-3 x m_{2} = -1

m_{2} = 1/3

so, the equation of BE will be

y – 3 = 1/3 (x – 3)

3y – 9 = x – 3

x – 3y + 6 = 0

Thus, the required line equation is x – 3y + 6 = 0.

**34. Find the equation of the right bisector of the line segment joining the points (1, 2) and (5, – 6).**

**Solution: **

The slope of the line joining the points (1, 2) and (5, -6) is

m_{1} = (-6 – 2)/ (5 – 1) = -8/4 = -2

Now, if m_{2} is the slope of the right bisector of the above line

Then,

m_{1} x m_{2} = -1

-2 x m_{2} = -1

m_{2} = ½

The mid-point of the line segment joining (1, 2) and (5, -6) will be

((1 + 5)/2, (2 – 6)/2) = (6/2, -4/2) = (3, -2)

So, equation of the line is

y + 2 = ½ (x – 3)

2y + 4 = x – 3

x – 2y – 7 = 0

Thus, the equation of the required right bisector is x – 2y – 7 = 0.

**35. 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’ if ( – 2, p) lies on it.**

**Solution: **

Given, co-ordinates of points A are (7, -3) and of B are (1, 9)

(i) The slope of AB (m) = (9 + 3)/ (1 – 7) = 12/ (-6) = -2

(ii) Let PQ be the perpendicular bisector of AB intersecting it at M

Now, the co-ordinates of M will be the mid-point of AB

Co-ordinates of M will be

= (7 + 1)/2, (-3 + 9)/2 = 8/2, 6/2

= (4, 3)

The slope of line PQ will be = -1/m = -1/ (-2) = ½

Thus, the equation of PQ is

y – 3 = ½ (x – 4)

2y – 6 = x – 4

x – 2y + 2 = 0

(iii) As point (-2, p) lies on the above line

The point will satisfy the line equation

-2 – 2p + 2 = 0

-2p = 0

p = 0

Thus, the value of p is 0.

**36. The points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.**

**Solution: **

Given, points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD

Slope of BD is given by

m_{1 }= (8 – 3)/ (6 – 1) = 5/5 = 1

We know that, the diagonal AC is a perpendicular bisector of diagonal BD

So, the slope of AC (m_{2}) will be

m_{1} x m_{2} = -1

1 x m_{2} = -1

m_{2} = -1

And, the co-ordinates of mid-point of BD and AC will be

((1 + 6)/2 , (3 + 8)/2) = (7/2, 11/2)

So, the equation of AC is

y – 11/2 = -1 (x – 7/2)

2y – 11 = -2x – 7

2x + 2y – 7 – 11 = 0 ⇒ 2x + 2y – 18 = 0

Thus, the equation of diagonal AC is x + y – 9 = 0.

**37. ABCD is a rhombus. The co-ordinates of A and C are (3, 6) and ( – 1, 2) respectively. Write down the equation of BD.**

**Solution: **

Given, ABCD is a rhombus and co-ordinates of A are (3, 6) and of C are (-1, 2)

Slope of AC (m_{1}) = (2 – 6)/ (-1 – 3) = -4/-4 = 1

We know that, the diagonals of a rhombus bisect each other at right angles.

So, the diagonal BD is perpendicular to diagonal AC

Let the slope of BD be m_{2}

Then, m_{1} x m_{2} = -1

m_{2} = -1/(m_{1})

= -1/ (1) = -1

Now, the co-ordinates of the mid-point of AC is given by

((3 – 1)/2, (6 + 2)/2) = (2/2, 8/2) = (1, 4)

So, the equation of BD will be

y – 4 = -1 (x – 1)

y – 4 = -x + 1

x + y = 5

Thus, the equation of BD is x + y = 5.

**38. Find the equation of the line passing through the intersection of the lines 4x + 3y = 1 and 5x + 4y = 2 and**

**(i) parallel to the line x + 2y – 5 = 0 **

**(ii) perpendicular to the x-axis.**

**Solution: **

Given, line equations:

4x + 3y = 1 … (1)

5x + 4y = 2 … (2)

On solving the above equation to find the point of intersection, we have

Multiplying (1) by 4 and (2) by 3

16 x + 12y = 4

15x + 12y = 6

On subtracting, we get

x = -2

Putting the value of x in (1), we have

4(-2) + 3y = 1

-8 + 3y = 1

3y = 1 + 8 = 9

y = 9/3 = 3

Hence, the point of intersection is (-2, 3).

(i) Given line, x + 2y – 5 = 0

2y = -x + 5

y = -(1/2) x + 5/2

Slope (m) = -1/2

A line parallel to this line will have the same slope m = -1/2

So, the equation of line having slope m and passing through (-2, 3) will be

y – 3 = (-1/2) (x + 2)

2y – 6 = -x – 2

x + 2y = 4

(ii) As any line perpendicular to x-axis will be parallel to y-axis.

So, the equation of line will be

x = -2 ⇒ x + 2 = 0

**39. (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 0 is the origin **

**(iii) In what ratio does the y-axis divide the line AB?**

**Solution: **

(i) Given, co-ordinates of the line joining A ( – 4, 1) and B (17, 10) and point P divides the line segment in the ratio 1 : 2

Let the co- ordinates of P be (x, y)

Then,

**40. Find the image of the point (1, 2) in the line x – 2y – 7 = 0.**

**Solution: **

Given line equation: x – 2y – 7 = 0 … (i)

Draw a perpendicular from point P (1, 2) on the line

Let P’ be the image of P and let its co-ordinates be (x, y)

The slope of the given line is given as,

2y = x – 7

y = (1/2) x – 7

Slope (m_{1}) = ½

Let the slope of line segment PP’ be m_{2}

As PP’ is perpendicular to the given line, product of slopes: m_{1} x m_{2} = -1

So, ½ x m_{2} = -1

m_{2} = -2

So, the equation of the line perpendicular to the given line and passing through P (1, 2) is

y – 2 = (-2) (x – 1)

y – 2 = -2x + 2

2x + y – 4 = 0 … (ii)

Let the intersection point of lines (i) and (ii) be taken as M.

Solving both the line equations, we have

Multiplying (ii) by 2 and adding with (i)

x – 2y – 7 = 0

4x + 2y – 8 = 0

———————

5x – 15 = 0

x = 15/5 = 3

Putting value of x in (i), we get

3 – 2y – 7 = 0

2y = -4

y = -4/2 = -2

So, the co-ordinates of M are (3, -2)

Hence, its seen that M should be the mid-point of the line segment PP’

(3, -2) = ((x + 1)/2, (y + 2)/2)

(x + 1)/2 = 3

x + 1 = 6

x = 6 – 1 = 5

And,

(y + 2)/2 = -2

y + 2 = -4

y = -4 – 2 = -6

Therefore, the co-ordinates of P’ are (5, -6).

**41. If the line x – 4y – 6 = 0 is the perpendicular bisector of the line segment PQ and the co-ordinates of P are (1, 3), find the co-ordinates of Q.**

**Solution: **

Given, line equation: x – 4y – 6 = 0 … (i)

Co-ordinates of P are (1, 3)

Let the co-ordinates of Q be (x , y)

Now, the slope of the given line is

4y = x – 6

y = (1/4) x – 6/4

slope (m) = ¼

So, the slope of PQ will be (-1/m) [As the product of slopes of perpendicular lines is -1]

Slope of PQ = -1/ (1/4) = -4

Now, the equation of line PQ will be

y – 3 = (-4) (x – 1)

y – 3 = -4x + 4

4x + y = 7 … (ii)

On solving equations (i) and (ii), we get the coordinates of M

Multiplying (ii) by 4 and adding with (i), we get

x – 4y – 6 = 0

16x + 4y = 28

——————

17x = 34

x = 34/17 = 2

Putting the value of x in (i)

2 – 4y – 6 = 0

-4 – 4y = 0

4y = -4

y = -1

So, the co-ordinates of M are (2, -1)

But, M is the mid-point of line segment PQ

(2, -1) = (x + 1)/2 , (y + 3)/2

(x + 1)/2 = 2

x + 1 = 4

x = 3

And,

(y + 3)/2 = -1

y + 3 = -2

y = -5

Thus, the co-ordinates of Q are (3, -5).

**42. OABC is a square, O is the origin and the points A and B are (3, 0) and (p, q). If OABC lies in the first quadrant, find the values of p and q. Also write down the equations of AB and BC.**

**Solution: **

Given, OABC is a square

Co-ordinates of A and B are (3, 0) and (p, q) respectively

By distance formula, we have

Chapter Test

**1. Find the equation of a line whose inclination is 60° and y-intercept is – 4.**

**Solution: **

Given, inclination = 60^{o} and y-intercept (c) = -4

So, slope (m) = tan 60^{o} = √3

Hence, the equation of the line is given by

y = mx + c

y = √3x – 4

**2. Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.**

**Solution: **

Given line equation: 3y + 2x = 12

3y = -2x + 12

y = (-2/3) x + 12/3

y = (-2/3) x + 4

Hence, gradient = -2/3 and the intercept on the y-axis is 4.

**3. If the equation of a line is y – √3x + 1, find its inclination.**

**Solution: **

Given line equation: y – √3x + 1

y = √3x – 1

Here, slope = √3

⇒ tan θ = √3

Θ = 60^{o}

Hence, the inclination of the line is 60^{o}.