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

**Exercise 12.1**

**1. Find the slope of a line
whose inclination is**

**(i) 45° **

**(ii) 30**

**°**

**Answer**

**(i)** tan 45° = 1

**(ii)** tan 30° =
1/√3

**2. Find the inclination of a
line whose gradient is **

**(i) 1 **

**(ii) √3 **

**(iii) 1/√3**

**Answer**

**(i)** tan θ = 1 ⇒ θ = 45°

**(ii)** tan
θ = √3 ⇒ θ = 60°

**(iii)** tan
θ = 1/√3 ⇒ θ = 30°

**3. Find the equation of a
straight line parallel 1 to x-axis is at a distance **

**(i) 2 units above it **

**(ii) 3 units below it. **

**Answer**

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

⇒ y = 2

⇒ y – 2 = 0

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

⇒ y = -
3

⇒ 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. **

**Answer**

**(i)** The equation of line parallel
to y-axis is at a distance of 3 units to the right is x = 3

⇒ x – 3
= 0

**(ii)** The equation of line parallel
to y-axis at a distance of 2 units to the left is x = - 2

⇒ x + 2
= 0

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

**Answer**

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 **

**Answer**

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

y = mx + c

where m is the slope and c is the
y-intercept

**(i)** y = mx + c

⇒ y = 3x
+ (-5)

⇒ y = 3x
– 5

**(ii)** y = mx + c

⇒ y =
-2/7.x + 3

⇒ 7y = -
21x + 21

⇒ 2x +
7y – 21 = 0

**(iii) **

**(iv)** Inclination = 30°

∴ Slope = tan 30° = 1/√3

∴ Equation
y = mx + c

⇒ y =
1/√3 x + 2

⇒ √3y =
x + 2√3

⇒ x -
√3y + 2√3 = 0

**7. Find the slope the 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 **

**Answer**

We know that in the equation

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

Now using this, we find,

**(i)** x – 2y – 1 = 0

⇒ x – 1
= 2y

⇒ 2y = x
– 1

⇒ y =
1/2 x – 1/2

Here, slope = 1/2 and y-intercept =
-1/2

**(ii)** 4x – 5y – 9 = 0

⇒ 4x – 9
= 5y

⇒ 5y =
4x – 9

⇒ y =
4/5 x – 9/5

Here, slope = 4/5 and intercept =
-9/5

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

⇒ 5y = - 3x – 7

⇒ y =
-3/5 x – 7/5

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

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

⇒ 4x +
3y = 12

⇒ 3y = -
4x + 12

⇒ y =
-4/3.x + 12/3

⇒ y =
-4/3 x + 4

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

**(v)** y – 3 = 0

⇒ y = 3

⇒ y = 0, x + 3

Here, slope = 0 and y-intercept = 3

**(vi) **x – 3 = 0

Here in this equation, slope cannot
be defined and 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 PQ makes with the positive
direction of x-axis. **

**Answer**

Equation of line PQ is 3y – 3x + 7
= 0

Writing in form of y = mx + c

3y = 3x – 7

⇒ y =
3x/3 – 7/3

⇒ y = x
– 7/3

**(i)** Here slope = l

**(ii)** ∴ Angle which makes PQ with x-axis
is Q.

But tan θ = 1

∴ θ = 45°

**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. **

**Answer**

Slope of
the line y = x + 1 after comparing it with y = mx + c, m = 1

⇒ tan θ = 1

⇒ θ = 45°

and slope
of line y = √3x – 1

m = √3

⇒ tan θ = √3

⇒ θ = 60°

Now in ∆
formed by the given two lines and x-axis.

Ext.
angle = Sum of interior opposite angle.

⇒ 60° = θ + 45°

⇒ θ = 60°
- 45°

= 15°

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

**Answer**

Equation of line is y/2 = x – p

It passes through the points (- 4,
4) and it will satisfy the equation

⇒ 4/2 =
- 4 – p

⇒ 2 = -
4 – p

⇒ p = -
4 – 2

⇒ p = -
6

Hence, p = - 6

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

**Answer**

Equation of line is y/2 = x – p

It passes through the points (- 4,
4) and it will satisfy the equation

⇒ 4/2 =
-4 – p

⇒ 2 = -4 – p

⇒ p = -4 – 2

⇒ p = -6

Hence, p = - 6

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

**Answer**

∵ Point (a, 2a) lies on the line

y/2 = 3x – 6

∴ this point will satisfy the equation

∴ 2a/2 = 3(a) – 6

⇒ a = 3a – 6

- 3a + a = - 6

⇒ - 2a = - 6

⇒ a = -6/-2

∴ 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. **

**Answer**

Equation of the line is y = mx + c

∴ It passes through the points (1, 4)

∴ 4 = m×1 + c

⇒ 4 = m + c

Hence, m + c = 4** …(i)**

Again it passes through the point (- 2, - 5)

∴ 5 = m(-2) + c

⇒ 5 = -2m + c

So, 2m – c = 5 **…(ii)**

Adding (i) and (iii)

3m = 9

⇒ m = 3

Substituting the value of m in (i)

3 + c = 4

⇒ c = 4 – 3 = 1

Hence, 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. **

**Answer**

∴ The line intersects y-axis making an intercept of -3

∴ The co-ordinates of point of intersect will be (0, - 3)

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

= (-3+5)/(0 – 2)

= 2/-2

= - 1

∴ Equation of the line will be,

y – y_{1} = m(x – x_{1})

⇒ y – (-5) = -1(x – 2)

⇒ y + 5 = -x + 2

⇒ x + y + 5 – 2 = 0

⇒ x + y + 3 = 0

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

**Answer**

Equation of the line will be

y – y_{1 }= m(x – x_{1})

y – 2 = 2/5(x + 1)

⇒ 5y – 10 = 2x + 2

So, 2x – 5y + 2 + 10 = 0

Hence 2x – 5y + 12 = 0

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

**Answer**

The equation of line whose slope is wand passes through a given point is y - y_{1} = m(x – x_{1})

Here m = tan 60° = √3 and point is (0, - 3)

∴ y + 3 = √3 (x – 0)

⇒ y + 3 = √3x

⇒ √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) **

**Answer**

m = (y_{2} – y_{1})/(x_{2} – x_{1})

Given

(i) (0, - 2), (3, 4)

(ii) (3, - 7), (- 1, 8)

**(i)** m = (4 + 2)/(3 – 0) = 6/3 = 2

∴ gradient = 2

**(ii)** m = (8 + 7)/(-1 – 3) = 15/-4

∴ 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. **

**Answer**

Co-ordinates of points E (0, 4) and F (3, 7) are given, then

**(i)** The gradient of EF

∴ gradient (m) = (y_{2} – y_{1})/(x_{2} – x_{1})

= (7 – 4)/(3 – 0)

= 3/3

= 1

**(ii)** Equation of line EF,

y – y_{1 }= m(x – x_{1})

⇒ y – 7 = 1(x – 3)

⇒ y – 7 = x – 3

⇒ x – y – 3 + 7 = 0

⇒ x – y + 4 = 0

**(iii)** Co-ordinates of point of intersection of EF and the x-axis will be y = 0,

Substitutes the value y in the above equation x – y + 4 = 0

⇒ x – 0 + 4 = 0 **(∵ y = 0)**

⇒ x = - 4

Hence co-ordinates are (- 4, 0)

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

**Answer**

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

And x = - 6

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

2x – 3y + 12 = 0

⇒ 2 × 0 – 3y + 12 = 0

⇒ -3y = - 12

⇒ y = -12/-3

= 4

**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. **

**Answer**

The two given points are P (5, 1), Q (1, -1).

∴ Slope of the line (m) = (y_{2} – y_{1})/(x_{2} – x_{1}) = (- 1 – 1)/(1 – 5)

= -2/-4

= 1/2

Equation of the line,

y – y_{1 }= m(x – x_{1})

⇒ y + 1 = 1/2(x–1)

⇒ 2y + 2 = x – 1

⇒ x – 2y – 1 – 2 = 0

⇒ x – 2y – 3 = 0

If point R (11, 4) be on it, then it will satisfy it.

Now substituting the value of x and y in

11 – 2×4 – 3

= 11 – 8 – 3

= 11 – 11

= 0

∴ R satisfies it.

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. **

**Answer**

Given that

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

Slope of AB = Slope of BC

⇒ (1 – 3)/(2 – a) = (a – 1)/(5 – 2)

⇒ -2/(2 – a) = (a – 1)/3

⇒ -6 = (a – 1)(2 – a) **(Cross-multiplication)**

⇒ -6 = 2a – a^{2 }– 2 + a

⇒ -6 = 3a – a^{2} – 2

⇒ a^{2} – 3a + 2 – 6 = 0

⇒ a^{2} – 3a – 4 = 0

⇒ a^{2} – 4a + a – 4 = 0

⇒ a(a – 4) + 1(a – 4) = 0

⇒ (a + 1)(a – 4) = 0

⇒ a = -1, or a = 4

a = - 1 **(∵ does not satisfy the equation)**

∴ a = 4

Slope of BC = (a–1)/(5–2)

= (4–1)/3

= 3/3

= 1 m

Equation of BC ; (y–1) = 1(x–2)

y – 1 = x – 2

⇒ x – y = -1 + 2

⇒ 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 (1/2, k). **

**Answer**

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

From your graph, we see 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. **

**Answer**

Given that

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

**(i) **

For the line AC

3.5 = (x + 4)/2

2 = (y + 7)/2

⇒ x + 4 = 7

⇒ y + 7 = 4

⇒ x = 7 – 4 = 3

⇒ y = 4 – 7 = - 3

x = 3, y = - 3

Thus, the coordinates 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 = 4x – 20

⇒ 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. **

**Answer**

AD is median

⇒ D is mid point of BC

∴ D is (7+1)/2, (8–10)/2}

i.e., (4, -1)

Slope of AD

m = (y_{2} – y_{1})/(x_{2} – x_{1})

= (5+1)/(3–4) = 6/-1 = - 6

∴ Equation of AD

y – y_{1} = m(x – x_{1})

⇒ y + 1 = - 6(x – 4)

⇒ y + 1 = - 6x + 24

⇒ y + 6x = - 1 + 24

⇒ 6x + y = 23

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

**Answer**

x–intercept = 4

∴ Co-ordinates of the point will be (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})

= (0 - 3)/(4 + 2)

= - 3/6

= - (1/2)

∴ Equation of the line will be y – y_{1} = m(x – x_{1})

⇒ y - 0 = - {1/2(x – 4)}

⇒ 2y = - x + 4

⇒ x + 2y = 4

or, x + 2y – 4 = 0

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

**Answer**

x–intercept = 6

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

y – intercept = - 4

⇒ c = - 4

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

Now m = (y_{2} – y_{1})/(x_{2} – x_{1})

= (-4–0)/(0–6)

= -4/-6

= 2/3

∴ Equation of line will be y = mx + c

⇒ y = 2/3 x + (-4) = 2/3 x – 4

⇒ 3y = 2x – 12

⇒ 2x – 3y = 12

**26. Write down the equation of the line whose gradient is 1/2 and which passes through P where P divides the line segment joining A (- 2, 6) and B (3, - 4) in the ratio 2 : 3.**

**Answer**

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

∴ 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

∴ Co-ordinates are (0, 2)

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

∴ Equation of the line 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. **

**Answer**

Line x – 2y – 11 = 0 passes through y-axis

x = 0,

Now substituting the value of x in the equation x – 2y – 11 = 0

∴ - 2y – 11 = 0

⇒ - 2y = 11

⇒ y = - (11/2)

∴ Co-ordinates of point will be (0, - 11/2)

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

m = (y_{2} – y_{1})/(x_{2} – x_{1}) = (- 11/2 – 4)/(0–1)

= (-19/2)/(-1)

= 19/2

And equation of the line 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. **

**Answer**

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

OA = OB given

∴ x = y

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

= (0 – y)/(x – 0) = - x/x = - 1 **(∵ x = y)**

∴ Equation of the line will be

y – y_{1} = m(x – x_{1})

⇒ y – 2 = -1×(x – 3)

⇒ y – 2 = -x + 3

⇒ x + y – 2 – 3 = 0

⇒ x + y – 5 = 0

⇒ x + y = 5

**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. **

**Answer**

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

**(i)** We need to find the co-ordinates of D

We know that the diagonals of a parallelogram bisect each other

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

∴ Mid-point of diagonal AC = {(3 + 3)/2, (6 + 2)/2}

= (3, 4)

And, mid-point of diagonal BD = {(5 + x)/2, (10 + y)/2}

Thus, we have

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

⇒ 5 + x = 6 and 10 + y = 8

⇒ x = 1 and y = - 2

∴ Co-ordinate of D = (1, - 2)

**(ii) **

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

(x – 3)/(3 – 1) = (y – 6)/(6 + 2)

⇒ (x – 3)/2 = (y – 6)/8

⇒ 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. **

**Answer**

Points A and B on x-axis and y-axis respectively

Let co-ordinates of A be (X, O) and of B be (O, Y)

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

Then, 2 = (X + O)/2 and – 3 = (O + Y)/2

⇒ x = 4, y = - 6

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

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

= (-6 – 0)/(0–4)

= -6/-4

= 3/2

**(iii)** Equation of AB will be y – y_{1} = m(x – x_{1})

⇒ y = (-3) = 3/2 (x – 2) **(∵ P lies on it)**

⇒ y + 3 = 3/2 (x – 2)

⇒ 2y + 6 = 3x – 6

⇒ 3x – 2y = 6 + 6

⇒ 3x – 2y = 12

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

**Answer**

The equations of sides of a rectangle whose equations 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.

Co-ordinates of A, B, C and D will be

(- 1, - 2), (2, - 2), (2, 6) and (- 1, 6) respectively.

AC and BD are its diagonals

(i) Slope of the diagonal AC = (y_{2} – y_{1})/(x_{2} – x_{1})

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

= 8/3

∴ Equation of AC will be

y – y_{1 }= m(x – x_{1})

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

⇒ 3y + 6 = 8x + 8

8x – 3y + 8 – 6 = 0

⇒ 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 **

**Answer**

5x + 7y = 3 **…(i)**

2x – 3y = 7 **…(ii)**

Multiply (i) by 3 and (ii) by 7,

15x + 21y = 9

14x – 21y = 49

Adding we get,

29x = 58

⇒ x = 58/29 = 2

Substituting the value of x in (i)

5 × 2 + 7y = 3

⇒ 10 + 7y = 3

⇒ 7y = 3 – 10

⇒ 7y = - 7

⇒ y = - 1

∴ Point of intersection of lines is (2, -1)

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

m = (y_{2} – y_{1})/(x_{2} – x_{1})

= (0 + 1)/(0 – 2)

= 1/2

Equation of line will be

y – y_{1} = m(x – x_{1})

⇒ y – 0 = - 1/2×(x – 0)

⇒ 2y = -x

⇒ 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. **

**Answer**

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

∴ Co-ordinates of A’ will be (3, 2)

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

∴ Co-ordinates of B will be (4, 3)

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

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

= -1/-1

= 1

Now tan θ= 1

∴ θ = 45°

Hence angle of inclination = 45°

**Exercise 12.2 **

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

**(i) Parallel **

**(ii) Perpendicular **

**(iii) neither nor perpendicular. **

**Answer**

Slope of line y = 3x – 5 = 3

And slope of line = 4x + 7

⇒ y = 2x + 7/2 = 2

∴ Slope of both lines are neither equal nor their product is -1.

∴ These line are neither parallel nor perpendicular.

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

**Answer**

In equation

6x + 5y – 7 = 0

⇒ 5y = -6x + 7

⇒ y = -(6/5)x + 2/5

∴ Slope (m) = -6/5 **…(i)**

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

⇒ 5y = - 2px – 1

⇒ y = -2/5 px – 1/5

∴ Slope (m) = -2/5 p **…(ii)**

∵ lines are parallel

m_{1} = m_{2}

From (i) and (ii) – (6/5) = -(2p/5)

⇒ p = -6/5 × (-5/2)

= 3

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

**Answer**

In equation 2x – by + 5 = 0

⇒ -by = -2x – 5

⇒ y = 2/b + 5/b

Slope (m) = 2/b

And in equation ax + 3y = 2

⇒ 3y = -ax + 2

⇒ y = -(a/3)×x + 2/3

∴ slope (m_{2}) = -(a/3)

∵ Lines are parallel

∴ m_{1} = m_{2}

⇒ 2/b = -(a/3)

⇒ - ab = 6

⇒ ab = -6

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

**Answer**

In equation y = x - p

⇒ y = 2x – 2p

Slope (m_{1}) = 2

In equation ax + 5 = 3y

⇒ y = a/3 ×x + 5/3

Slope (m_{2}) = a/3

∵ Lines are parallel

∴ m_{1} = m_{2}

a/3 = 2

⇒ a = 6

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

**Answer**

Gradient m_{1} of the line y = 3x + 7 is 3

2y + px = 3

⇒ y = (-px)/2 + 3/2

Gradient m^{2} of this line is – (p/2)

Since, the given lines are perpendicular to each other.

∴ m_{1} × m_{2} = - 1

⇒ 3 × (-p/2) = - 1

⇒ p = 2/3

**6. If the straight lines 3x – 5y + 4 = 0 and 4x – 2y + 5 = 0 are perpendicular to each other. **

**Answer**

Given

In equation, 3x – 5y + 4 = 0

⇒ 5y = kx + 4

⇒ y = k/5 + 4/5

∴ Slope (m_{1}) = k/5

And in equation, 4x – 2y + 5 = 0

⇒ 2y = 4x + 5

⇒ y = 2x + 5/2

Slope (m_{2}) = 2

∵ Lines are perpendicular to each other

∴ m_{1}m_{2} = - 1

k/5 × 2 = - 1

⇒ k = (-1 × 5)/2

= -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. **

**Answer**

Given,

In the equation 3x + by + 5 = 0

by = -3x – 5

⇒ y = -3/b ×x – 5/b

Slope (m_{1}) = -3/b

And in the equation ax - 5y + 7 = 0

⇒ 5y = ax + 7

⇒ y = a/5 + 7/5

∴ Slope (m_{2}) = a/5

∵ Lines are perpendicular to each other

∴ m_{1}m_{2} = - 1

⇒ -3/b × a/5 = - 1

⇒ - 3a/5b = - 1

⇒ - 3a = - 5b

⇒ 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).**

**Answer**

Slope of the line passing through the points (- 2, 3) and (4, 1) = (y_{2} – y_{1})/(x_{2} – x_{1})

= (1 – 3)/(4 + 2)

= -2/6

= -1/3

Slope of line 3x = y + 1

∵ m_{1} × m_{2} = -1/3 × 3 = - 1

∴ These lines are perpendicular to each other

Co-ordinates of mid-point of line joining the points (-2, 3) and (4, 1) will be {(-2+4)/2, (3+1)/2} or (2/2, 4/2) or (1, 2).

If mid-point (1, 2) lies on the line 3x = y + 1 then it will satisfy it.

Now substituting the value of x and y is 3x = y + 1

⇒ 3 ×1 = 2 + 1

⇒ 3 = 3 which is true.

Hence the line 3x = y + 1 bisects the line joining the points (-2, 3), (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. **

**Answer**

Gradient (m_{1}) of the line passing through the points A (- 2, 3) and B (4, b)

= (b – 3)/(4 + 2)

= (b - 3)/6

Gradient (m_{2}) of the line 2x – 4y = 5

Or y = x/2 – 5/2 is 1/2

Since, the lines are perpendicular to each other,

∴ m_{1} × m_{2} = - 1

(b – 3)/6 × 1/2 = - 1

⇒ (b – 3)/12 = - 1

⇒ b – 3 = - 12

⇒ b = - 9

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

**Answer**

In the line 3x + y = 4 **…(i)**

⇒ y = -3x + 4

Slope (m_{1}) = - 3

In the line x – ay + 7 = 0 **…(ii)**

⇒ ay = x + 7

⇒ y = (1/a)×x + 7/a

Slope (m_{2}) = 1/a

And in the line bx + 2y + 5 = 0** …(iii)**

⇒ 2y = -bx – 5

⇒ y = (-b/2) ×x – 5/2

∴ Slope (m_{3}) = - b/2

∵ These are the consecutive three sides of a rectangle.

∴ (i) and (ii) are perpendicular to each other

∴ m_{1}m_{2} = - 1

⇒ -3 × 1/a = - 1

⇒ -3 = - a

⇒ a = 3

And (i) and (ii) are parallel to each other

∴ m_{1} = m_{3}

⇒ -3 = (- b)/2

⇒ -b = - 6

⇒ b = 6

Hence a = 3, b = 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. **

**Answer**

In the given line 2x – 3y – 7 = 0

⇒ 3y = 2x – 7

⇒ y = (2/3)×x – 7/3

Hence slope (m_{1}) = 2/3

∴ Equation of the line parallel to the given line will be

y – y_{1} = m(x – x_{1})

∵ it passes through (0, 4), then

y – 4 = 2/3(x – 0)

⇒ 3y – 12 = 2x

⇒ 2x – 3y + 12 = 0 **…(i)**

Now let it intersect x-axis at (x, y)

∴ y = 0

Substitute the value of y in (i)

2x – 3×0 + 12 = 0

⇒ 2x = - 12

x = - 6

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

**Answer**

In the line 2x + 5y + 7 = 0

⇒ 5y = -2x – 7

⇒ y = (-2/5)×x – 7/5

Here slope (m_{1}) = -(2/5)

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

∴ m_{1}m_{2} = - 1

⇒ -(2/5)m_{2 }= - 1

∴ m_{2 }= -1 × -5/2

= 5/2

∵ It makes y-intercept –3 units

∴ The point where it passes = (0, -3)

∴ Equations of the new line,

y – y_{1} = m(x - x_{1})

⇒ y – (-3) = 5/2×(x – 0)

⇒ y + 3 = (5/2)×x

⇒ 2y + 6 = 5x

⇒ 5x – 6y – 6 = 0

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

**Answer**

In the given line 3x – 4y + 12 = 0

⇒ 4y = 3x + 12

⇒ y = (3/4)x + 3

Here slope (m_{1}) = 3/4

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

∴ m_{1}m_{2 }= - 1

⇒ 3/4m_{2} = - 1

m_{2} = -4/3

y-intercept in the equation

2x – y + 5 = 0

⇒ 2×0 – y + 5 = 0

⇒ y = 5

∴ The equation of the line passing through (0, 5) will be

y – y_{1} = m(x – x_{1})

⇒ y – 5 = -4/3 ×(x – 0)

⇒ 3y – 15 = - 4x

⇒ 4x + 3y – 15 = 0

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

**Answer**

In the given line 3x – 2y = - 4

⇒ 2y = 3x + 4

⇒ y = 3/2×x + 2

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

∴ Slope of the line parallel to the given line = 3/2 and passes through (0, 3)

∴ Equation of the line will be y – y_{1 }= m(x – x_{1})

⇒ y – 3 = 3/2 × (x – 0)

⇒ 2y – 6 = 3x

⇒ 3x – 2y + 6 = 0

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

**Answer**

In the given equation 3x + 5y + 15 = 0

⇒ 5y = - 3x – 15

⇒ y = -3/5 ×(x – 3)

How slope (m_{1}) = -3/5

∴ Slope of the line parallel to the given line = -3/5 and passes through the point (0, 4)

∴ 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). **

**Answer**

In the given line y = 3x – 5

Here slope (m_{1}) = 3

Substituting x = 0, then y = - 5

y- intercept = - 5

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

Equation of the line will be

y – y_{1} = m(x – x_{1})

⇒ y – 5 = 3(x – 0)

⇒ y – 5 = 3x

⇒ 3x – y + 5 = 0

⇒ y = 3x + 5

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

**Answer**

In the given line 3x + 8y = 12

⇒ 8y = - 3x + 12

⇒ y = -3/8 ×x + 12/8

Here slope (m_{1}) = -3/8

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

∴ m_{1}m_{2} = -1

⇒ -3/8 × m_{2} = -1

m_{2} = 8/3

∴ Equation of the line where slope is 8/3 and passes 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

⇒ 8x – 3y + 8 – 6 = 0

⇒ 8x – 3y + 2 = 0

**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. **

**Answer**

**(i)** In the line 4x – 3y + 12 = 0 **…(i)**

3y = 4x + 12

⇒ y = 4/3 ×x + 4

Here slope (m_{1}) = 4/3

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

∴ m_{1}m_{2} = - 1

⇒ 4/3 ×m_{2} = - 1

⇒ m_{2} = -3/4

Let the point on x-axis be A (x, 0)

∴ Substituting the value of y in (i)

4x – 3×0 + 12 = 0

⇒ 4x + 12 = 0

⇒ 4x = -12

⇒ x = -3

∴ Co-ordinates of A will be (- 3, 0)

**(ii)** Equation of the line perpendicular to the given line passing through A will be.

y – y_{1 }= m(x – x_{1})

⇒ y – 0 = -3/4 ×(x + 3)

⇒ 4y = - 3x – 9

⇒ 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). **

**Answer**

The given line 2x + 5y – 7 = 0

5y = - 2x + 7

⇒ y = -2/5 ×x + 7/5

Here slope (m_{1}) = -2/5

∴ Slope of the line parallel to the given line will be –2/5.

Co-ordinates of the mid-point joining the points (2, 7) and (-4, 1) will be = {(2–4)/2, (7+1)/2} or (-2/2, 8/2) or (-1, 4)

∴ 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 – 20 + 2 = 0

⇒ 2x + 5y – 18 = 0

**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), (2, 2). **

**Answer**

In the given line 3x + 2y – 8 = 0

⇒ 2y = -3x + 8

⇒ y = -3/2 ×x + 4

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

Co-ordinates of the mid-point of the line segment joining the points (5, -2) and (2, 2) will be {(5+2)/2, (-2+2)/2} or (7/2, 0)

And let the slope of the line perpendicular to the given line be = m_{2 }

∴ m_{1}m_{2 }= - 1

⇒ -3/2 m_{2} = - 1

⇒ m_{2 }= 2/3

∴ Equations of the line perpendicular to the given line and passing through (7/2, 0) will be.

y – y_{1} = m(x – x_{1})

⇒ y – 0 = 2/3(x – 7/2)

⇒ 3y = 2x – 7

⇒ 2x – 3y – 7 = 0

**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. **

**Answer**

Slope (m_{1}) of line by joining the points A (1, 3), B (3, - 1) = (y_{2} – y_{1})/(x_{2} – x_{1})

∴ m_{1} = (-1–3)/(3–1) = -4/2 = - 2

Slope (m_{2}) of the line joining the points B

(3, -1) and C (-5, -5) = (y_{2}–y_{1})/(x_{2}–x_{1})

⇒ m_{2} = (-5+1)/(-5–3)

= (-4/-8)

= 1/2

∴ m_{1} × m_{2} = -2 × 1/2 = -1

∴ Lines AB and BC are perpendicular to each other.

Hence ∆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). **

**Answer**

Slope of the line joining the points (0, - 2) and (4, 5) = (y_{2 }– y_{1})/(x_{2} – x_{1})

= (5+2)/(4–0)

= 7/4

Slope of the line parallel to it passing through (-1, 3) = 7/4

And equation of the line

y – y_{1} = m(x – x_{1})

⇒ y – 3 = 7/4 ×(x + 1)

⇒ 4y - 12 = 7x + 7

⇒ 7x – 4y + 7 + 12 = 0

⇒ 7x – 4y + 19 = 0

**28. A (- 1, 3), B (4, 2) , C (3, - 2) 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. **

**Answer**

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

**(i)** Coordinates of centroid G = {(x_{1 }+ x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3}

= {(-1 + 4 + 3)/3, (3 + 2 – 2)/3}

= (6/3, 3/3)

= (2, 1)

So, the coordinates are (2, 1)

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

= -5/4

∴ Slope of the required line (m) = -5/4

Let the equation of the line through G, be y – y_{1} = m(x – x_{1})

⇒ y – 1 = -5/4 ×(x–2)

⇒ 4y – 4 = -5x + 10

⇒ 5x + 4y – 14 = 0 which is the required line.

**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. **

**Answer**

**(i)** Here θ = 45°

So, slope of the line = tan θ

= tan 45°

= 1

**(ii)** Equation of the line through P and Q is y –3 = 1(x – 5)

⇒ y – x + 2 = 0

**(iii)** Let the coordinates of Q be (0, y)

Then m = (y_{2 }– y_{1})/(x_{2 }– x_{1})

⇒ 1 = (3 – y)/(5 – 0)

⇒ 5 = 3 – y

⇒ y = -2

So, coordinates 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. **

**Answer**

From the given figure, it is clear that co-ordinates of A are (2, 3) of B are (- 1, 2) and of C are (3, 0).

_{2}- y

_{1})/(x

_{2}– x

_{1})

= (0 – 2)/{3 – (-1)}

= -2/(3+1)

= -2/4

= -1/2

∴ Slope of line parallel to BC = -1/2

∵ It passes through A (2, 3)

∴ Its equations will be, y – y_{1} = m(x – x_{1})

⇒ y – 3 = -1/2 ×(x – 2)

⇒ 2y – 6 = -x + 2

⇒ x + 2y = 2 + 6

⇒ x + 2y = 8

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

**Answer**

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

= (y_{2} – y_{1})/(x_{2} – x_{1})

= (1 – 2)/(9 + 3)

= -1/12

Let slope of the line perpendicular to the line = m_{2}

∴ m_{1}m_{2} = -1

⇒ -1/12 × m_{2} = -1

⇒ m_{2} = -1 × (-12/1)

= 12

∴ Equation of the line passing through (0, -3) and of slope m_{2} = 12

y - y_{1 }= m(x – x_{1})

⇒ y + 3 = 12(x - 0)

⇒ y + 3 = 12x

⇒ 12x – y – 3 = 0

**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.**

**Answer**

Vertices of ∆ABC are A (10, 4), B (4, - 9) and C (-2, -1)

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

= (-1 + 9)/(-2 – 4)

= 8/(-6)

= -4/3

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

∴ m_{1}m_{2} = -1

⇒ -4/3 × m_{2} = -1

⇒ m_{2} = -1 ×(-3/4) = 3/4

∴ Equation of the line will be,

y – y_{1} = m(x – x_{1})

⇒ y – 4 = 3/4 ×(x – 10)

⇒ 4y – 16 = 3x – 30

⇒ 3x – 4y + 16 – 30 = 0

⇒ 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 **

**Answer**

(i) D is the mid-point of BC

Co-ordinates of D will {(3 – 1)/2, (3 + 5)/2} or {(2/2, 8/2)} or (1, 4)

∴ Slope of median AD

(m) = (y_{2} – y_{1})/(x_{2} – x_{1})

= (4 + 4)/(1 – 2)

= 8/-1

= -8

Then equation of AD will be,

y – y_{1 }= m(x – x_{1})

⇒ y – 4 = -8 (x – 1)

⇒ y – 4 = -8x + 8

⇒ 8x + y – 4 – 8 = 0

⇒ 8x + y – 12 = 0

**(ii)** BE is the altitude from B to AC

∴ Slope of AC (m_{1}) = (y_{2} – y_{1})/(x_{2 }- x_{1})

= (5 + 4)/(- 1 – 2)

= 9/-3

= -3

Let slope of BE = m_{2}

But m_{1}m_{2} = -1

⇒ - 3×m_{2} = -1

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

∴ Equation of BE will be,

y – y_{1} = m(x – x_{1})

⇒ y – 3 = 1/3 ×(x – 3)

⇒ 3y – 9 = x – 3

⇒ x – 3y – 3 + 9 = 0

⇒ x – 3y + 6 = 0

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

**Answer**

Slope of the line joining the points (1, 2) and (5, 6)

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

= (-6 – 2)/(5 – 1)

= -8/4

= -2

Let m_{2} be the right bisector of the line

∴ m_{1}m_{2} = -1

⇒ -2 × m_{2} = -1

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

mid-point of the line segment joining (1, 2) and (5, -6) will be {(1 + 5)/2, (2 – 6)/2} or (6/2, -4/2) or (3, -2)

∴ Equation of line, the right bisector will be y – y_{1} = m(x – x_{1})

⇒ y + 2 = 1/2 ×(x – 3)

⇒ 2y + 4 = x – 3

⇒ x – 2y – 3 – 4 = 0

⇒ 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. **

**Answer**

Coordinates of A are (7, -3), of B = (1, 9)

**(i)** ∴ Slope (m) = (y_{2} – y_{1})/(x_{2 }– x_{1})

= {9 – (-3)}/(1 – 7)

= (9+3)/(1 – 7)

= 12/-6

= -2

**(ii)** Let PQ is the perpendicular bisector of AB intersecting it at M.

∴ Co-ordinates of M will be = (x_{1} + x_{2})/2, (y_{1} + y_{2})/2

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

= 8/2, 6/2 or (4, 3)

∴ Slope of PQ = 1/2 (m_{1}, m_{2} = -1)

∴ Equation of PQ = y – y_{1} = m(x – x_{1})

⇒ y – 3 = 1/2 ×(x – 4)

⇒ 2y – 6 = x – 4

⇒ x – 2y + 6 – 4 = 0

⇒ x – 2y + 2 = 0

**(iii)** ∵ Point (-2, p) lies on it

∴ - 2 – 2p + 2 = 0

⇒ -2p + 0 = 0

⇒ -2p = 0

∴ p = 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. **

**Answer**

Slope of BD (m_{1}) = (y_{2} – y_{1})/(x_{2 }– x_{1})

= (8 – 3)/(6 – 1)

= 5/5

= 1

Diagonal AC is perpendicular bisector of diagonal BD

∴ Slope of AC = -1 (∵ m_{1}m_{2} = -1)

And co-ordinates of midpoint of BD will be {(1 + 6)/2, (3 + 8)/2} or (7/2, 11/2)

∴ Equation of AC,

y – y_{1} = m(x – x_{1})

⇒ y – 11/2 = -1 ×(x – 7/2)

⇒ y – 11/2 = -x + 7/2

⇒ 2y – 11 = -2x + 7

⇒ 2x + 2y – 11 – 7 = 0

⇒ 2x + 2y – 18 = 0

Or 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. **

**Answer**

Co-ordinates of A (3, 6), C (-1, 2)

Slope of AC (m_{1}) = (y_{2} – y_{1})/(x_{2} – x_{1})

= (2 – 6)/(-1 – 3)

= -4/-4

= 1

∴ Slope of BD = - 1 (∵ m_{1}m_{2} = -1)

And co-ordinates of mid point of AC will be {(3 – 1)/2, (6 + 2)/2} or (2/2, 8/2) or (1, 4)

∴ Equation of BD will be,

y – y_{1} = m(x – x_{1})

⇒ y – 4 = -1(x – 1)

⇒ y – 4 = -x + 1

⇒ x + y – 4 – 1 = 0

⇒ x + y – 5 = 0

**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. **

**Answer**

4x + 3y = 1 **…(i)**

5x + 4y = 2 **…(ii)**

Multiplying (i) by 4 and (ii) by 3

16x + 12y = 4

15x + 12y = 6

Subtracting (i) from (ii),

x = -2

Substituting the value of x in (i)

4(-2) + 3y = 1

⇒ -8 + 3y = 1

⇒ 3y = 1+8 = 9

⇒ y = 9/3 = 3

∴ Point of intersection = (-2, 3)

**(i)** In the line x + 2y – 5 = 0

⇒ 2y = – x + 5

⇒ y = -1/2 x + 5/2

∵ Slope (m_{1}) = -1/2

∴ Slope of its parallel line = -1/2

And equation of the parallel line y – y_{1} = m(x – x_{1})

⇒ y – 3 = -1/2 ×(x + 2)

⇒ 2y – 6 = -x – 2

⇒ x + 2y – 6 + 2 = 0

⇒ x + 2y – 4 = 0

**(ii)** ∵ Any line perpendicular to x-axis will be parallel to y – axis.

∴ Equation of the line will be

x = a

i.e., 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? **

**Answer**

**(i)** Co-ordinate A (-4, 1) and B (17, 10) P divides it in the ratio of 1 : 2

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

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

= {1×17 + 2×(-4)}/(1+2)

= (17 – 8)/3

= 9/3

= 3

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

= (1×10 + 2×1)/(1 + 2)

= (10+2)/3

= 12/3

= 4

∴ Co-ordinates of P will be (3, 4)

**(ii)** O is the origin

∴ Distance between O and P

**(iii)** Let y-axis divides AB in the ratio of m_{1} : m_{2}

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

⇒ 0 = {m_{1}×17 + m_{2}×(-4)}/(m_{1}+m_{2})

⇒ 17m_{1} – 4m_{2} = 0

⇒ 17m_{1} = 4m_{2}

⇒ m_{1}/m_{2} = 4/17

⇒ m_{1} : m_{2} = 4 : 17

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

**Answer**

Draw a perpendicular from the point P (1, 2) on the line, x – 2y – 7 = 0

Let P’ is the image of P and let its coordinates are (𝝰, 𝝱) slope of line x – 2y – 7 = 0

⇒ 2y = x – 7

⇒ y = 1/2 ×x – 7/2 is 1/2

∴ Slope of PP’ = -2 **(∵ m _{1}m_{2} = -1)**

∴ Equation of PP’

y – y_{1 }= m(x – x_{1}) = y – 2 = -2×(x – 1)

⇒ y – 2 = -2x + 2

⇒ 2x + y = 2 + 2

⇒ 2x + y = 4

∵ P’ (𝝰, 𝝱) lies on it

∴ 2𝝰 + 𝝱 = 4 **…(i)**

∵ P’ is the image of P in the line x – 2y – 7 = 0

∴ the lines bisects PP’ at M.

Or M is the mid-point of PP’

∴ Co-ordinates of M will be {(1+𝝰)/2, (2+𝝱)/2}

∵ M lies on the given line x – 2y – 7 = 0

∴ Substituting the value of x, y

(1+𝝰)/2 – {2(2 + 𝝱)/2} - 7 = 0

⇒ (1+𝝰)/2 – (2 + 𝝱) – 7 = 0

⇒ 1 + 𝝰 – 4 - 2𝝱 - 14 = 0

⇒ 𝝰 - 2𝝱 = 4 + 14 – 1 = 17 **…(ii)**

⇒ 𝝰 = 17 + 2𝝱

Substituting the value of 𝝰 in (i)

2×(17 + 2𝝱) + 𝝱 = 4

34 + 4𝝱 + 𝝱 = 4

⇒ 5𝝱 = 4 – 34 = - 30

𝝱 = -30/5

= -6

Substituting the value of 𝝱 in (i)

2𝝰 – 6 = 4

⇒ 2𝝰 = 4 + 6 = 10

𝝰 = 10/2 = 5

∴ Co-ordinates of P’ will be (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. **

**Answer**

Let the co-ordinates of Q be (𝝰, 𝝱) and let the line x – 4y – 6 = 0 is the

Perpendicular bisector of PQ and it intersects the line at M.

M is the mid point of PQ

Now slope of line x – 4y – 6 = 0

⇒ 4y = x – 6

⇒ y = 1/4 ×x – 6/4 is 1/4

∴ Slope of PQ = -4 **(∵m _{1}m_{2 }= -1)**

And equation of line PQ

y – y_{1} = m(x – x_{1})

⇒ y – 3 = -4 (x – 1)

⇒ y – 3 = -4x + 4

⇒ 4x + y – 3 – 4 = 0

⇒ 4x + y – 7 = 0

⇒ 4x + y = 7

∵ Q (𝝰, 𝝱) lies on it.

∴ 4𝝰 + 𝝱 = 7 **…(i)**

Now co-ordinates of M will be {(1 + 𝝰)/2, (3 + 𝝱)/2}

∵ M lies on the line x – 4y – 6 = 0

∴ {(1 + 𝝰)/2 – 4(3+𝝱)/2} – 6 = 0

⇒ 1 + 𝝰 – 4(3 + 𝝱) – 12 = 0

⇒ 1 + 𝝰 - 12 - 4𝝱 – 12 = 0

⇒ 𝝰 - 4𝝱 = 24 – 1 = 23 **...(ii)**

Multiply (i) by 4 and (ii) by 1

16𝝰 + 4𝝱 = 28

𝝰 - 4𝝱 = 23

Adding we get,

17𝝰 = 51

⇒ 𝝰 = 51/17 = 3

Substituting the value of 𝝰 in (i)

4×3 + 𝝱 = 7

⇒ 𝝱 = 7 – 12 = -5

∴ Co-ordinates of Q will be (3, -5).

**Question 42: QABC 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. **

**Answer**

^{2}= OA

^{2 }+ AB

^{2}

^{2 }+ q

^{2}= 9 + (3 – p)

^{2}+ q

^{2}

⇒ p^{2} + q^{2} = 9 + 9 + p^{2 }– 6p + q^{2 }

6p = 18

⇒ p = 18/6 = 3

Substituting the value of p in (i)

(3)^{2} + q^{2 } - 6(3) = 0

⇒ 9 + q^{2} – 18 = 0

⇒ q^{2} – 9 = 0

⇒ q^{2} = 9

⇒ q = 3

∴ p = 3, q = 3

∵ Equation AB will be x = 3

⇒ x – 3 = 0

**Multiple Choice Questions**

**Choose the correct answer from the given four options : **

**1. The slope of a line parallel to y-axis is **

**(a) 0 **

**(b) 1 **

**(c) –1 **

**(d) not defined **

**Answer**

(d) not defined

Slope of a line parallel to y-axis is not defined.

**2. The slope of a line which makes an angle of 30° with the positive direction of x-axis is **

**(a) 1 **

**(b) 1/√3**

**(c) √3 **

**(d) –1/√3**

**Answer**

(b) 1/√3

Slope of a line which makes an angle of 30° with positive direction of x-axis = tan 30° =

**3. The slope of the line passing through the points (0, -4) and (-6, 2) is **

**(a) 0 **

**(b) 1 **

**(c) –1 **

**(d) 6 **

**Answer**

(c) -1

Slope of the line passing through the points (0, -4) and (-6, 2)

(y_{2} – y_{1})/(x_{2} – x_{1}) = (2 + 4)/(- 6 – 0)

= 6/-6

= -1

**4. The slope of the line passing through the points (3, - 2) and (- 7, - 2) is **

**(a) 0 **

**(b) 1 **

**(c) –1/10**

**(d) not defined. **

**Answer**

(a) 0

Slope of the line passing through the points (3, -2) and (-7, 2)

(y_{2} – y_{1})/(x_{2} – x_{1}) = (-2+2)/(-7–3)

= (0/-10)

= 0

**5. The slope of the line passing through the points (3, -2) and (3, -4) is **

**(a) –2 **

**(b) 0 **

**(c) 1 **

**(d) not defined **

**Answer**

(d) not defined

Slope of the line passing through the points (3, -2) and (3, -4)

(y_{2} – y_{2})/(x_{2} – x_{1}) = (-4+2)/(3 – 3)

= -2/0

**6. The inclination of the line y = √3x – 5 is **

**(a) 30°**

**(b) 60°**

**(c) 45°**

**(d) 0°**

**Answer**

(b) 60°

The inclination of the line y = √3x – 5 is √3 = tan 60°

= 60°

**7. If the slope of the line passing through the points (2, 5) and (k, 3) is 2, then the value of k is **

**(a) –2**

**(b) –1**

**(c) 1 **

**(d) 2 **

**Answer**

(c) 1

Slope of the line passing through (2, 5) and (k, 3) is 2, then

m = (y_{2} – y_{1})/(x_{2} – x_{1})

⇒ 2 = (3 – 5)/(k – 2)

⇒ 2 = (-2)/(k – 2)

⇒ 2k – 4 = -2

⇒ 2k = 4 – 2

= 2

⇒ k = 2/2 = 1

**8. The slope of a line parallel to the line passing through the points (0, 6) and (7, 3) is **

**(a) –7/3**

**(b) -3/7**

**(c) –3**

**(d) 3**

**Answer**

(b) -3/7

Slope of the line parallel to the line passing through (0, 6) and (7, 3)

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

= (3 – 6)/(7 – 0)

= -3/7

**9. The slope of a line perpendicular to the line passing through the points (2, 5) and (- 3, 6) is **

**(a) –1/5**

**(b) 1/5 **

**(c) –5 **

**(d) 5 **

**Answer**

(d) 5

Slope of the line joining the points (2, 5), (-3, 6)

= (y_{2 }– y_{1})/(x_{2 }– x_{1})

= (6 – 5)/(-3 – 2)

= 1/-5 = -1/5

∴ Slope of the line perpendicular to this line = 5

**10. The slope of a line parallel to the line 2x + 3y – 7 = 0 is **

**(a) –2/3**

**(b) 2/3 **

**(c) –3/2**

**(d) 3/2 **

**Answer**

(a) -2/3

The slope of a line parallel to the line 2x + 3y – 7 = 0

Slope of the line

3y = - 2x + 7

⇒ y = -2/3 + 7/3

= -2/3

**11. The slope of a line perpendicular to the line 3x = 4y + 11 is **

**(a) 3/4 **

**(b) –3/4**

**(c) 4/3 **

**(d) –4/3**

**Answer**

(d) -4/3

Slope of a line perpendicular to the line 3x = 4y + 11 is

⇒ 4y = 3x – 11

⇒ y = 3/4 x - 11/4

Slope = 3/4

∴ Slope of the line perpendicular to this line = -4/3 (∵ m × n = -1)

**12. If the line 2x + 3y = 5 and kx – 6y = 7 are parallel, then the value of k is **

**(a) 4 **

**(b) –4 **

**(c) 1/4 **

**(d) – 1/4**

**Answer**

(b) -4

Lines 2x + 3y = 5 and kx – 6y = 7 are parallel

Slope of 2x + 3y = 5

Slope of kx – 6y = 7

⇒ 3y – 2x + 5

⇒ y = -2/3 ×x + 5/3

Slope of 2x + 3y = 5 is -2/3

And slope of kx – 6y = 7

6y = kx – 7

⇒ y = (k/6)×x – 7/6

∴ Slope = k/6

Since both lines are parallel

∴ -2/3 = k/6

⇒ k = (-2 × 6)/3

= -4

**13. If the line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0 are perpendicular to each other, then the value of k is **

**(a) 3/2 **

**(b) –3/2**

**(c) 2/3 **

**(d) –2/3**

**Answer**

(a) 3/2

Line 3x – 4y + 7 = 0 and 2x + ky + 5 = 0 are perpendicular to each other

∴ Product of their slopes = (m_{1}×m_{2}) = -1

Slope of 3x – 4y + 7 = 0

⇒ 4y = 3x + 7

⇒ y = 3/4 x + 7/4

Slope (m_{1}) = 3/4

And slope of 2x + ky + 5 = 0

ky = -2x – 5

y = (-2/k) ×x – 5/k

∴ Slope (m_{2}) = -2/k

Since the given lines are perpendicular to each other

∴ 3/4 × -2/k = -1

⇒ -6/4k = -1

⇒ -k = -6/4

⇒ k = 3/2

**Chapter Test**

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

**Answer**

Angle of inclination = 60°

Slope = tan θ

= tan 60°

= √3

Equation of the line will be,

y = mx + c

= √3x + (-4)

⇒ y - √3x – 4

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

**Answer**

Slope of the line 3y + 2x = 12

⇒ 3y = 12 – 2x

⇒ 3y = - 2x + 12

⇒ y = (-2/3)×x + 4

∴ Slope = -2/3 and y-intercept = 4

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

**Answer**

In the line

y = √3x + 1

Slope = √3

⇒ tan θ = √3

⇒ θ = 60° **(∵ tan 60° = √3)**

**4. If the line y = mx + c passes through the points (2, -4) and (-3, 1), determine the values of m and c. **

**Answer**

The equation of line mx + c

∵ it passes through (2, -4) and (-3, 1)

Now substituting the value of these points –4 = 2m + c **…(i)**

And 1 = -3m + c **…(ii)**

Substracting we get,

-5 = 5m

⇒ m = -5/5 = -1

Substituting the value of m in (i)

-4 = 2(-1) + c

⇒ -4 = -2 + c

c = -4 + 2 = -2

∴ m = -1, c = -2

**5. If the point (1, 4), (3, -2) and (p, - 5) lie on a straight line, find the value of p. **

**Answer**

Let the points to be A (1, 4), B (3, -2) and C (p, -5) are collinear and let B (3, -2)

Divides AC in the ratio of m_{1} : m_{2 }

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

⇒ 3 = (m_{1}p + m_{2}×1)/(m_{1 }+ m_{2})

3m_{1 }+ 3m_{2} = m_{1}p + m_{2}

⇒ 3m_{1} – m_{1}p = m_{2} – 3m_{2}

⇒ m_{1}(3 – p) = -2 m_{2 }

⇒ m_{1}/m_{2} = - 2/(3 – p) **…(i)**

and –2 = {m_{1}(-5) + m_{2}×4}/(m_{1} + m_{2})

⇒ -2m_{1 }– 2m_{2 }= - 5m_{1} + 4m_{2}

⇒ -2m_{1} + 5m_{1} = 4 m_{2} + 2m_{2}

⇒ 3m_{1} = 6 m_{2}

⇒ m_{1}/m_{2} = 6/3 = 2 **…(ii)**

From (i) and (ii)

-2/(3–p) = 2

⇒ -2 = 6 – 2p

⇒ 2p = 6 + 2 = 8

⇒ p = 8/2 = 4

**6. Find the inclination of the line joining the points P (4, 0) and Q (7, 3).**

**Answer**

Slope of the line joining the points P (4, 0) and Q (7, 3)

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

∴ tan θ = 1

⇒ θ = 45° **(∵ tan 45° = 1)**

Hence inclination of line = 45°

**7. Find the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x – 2y = 5 and having y-intercept equal to –3/7. **

**Answer**

Equation of lines are

2x + y = 5 **…(i)**

x – 2y = 5 **...(ii)**

Multiply (i) by 2 and (ii) by 1, we get

4x + 2y = 10

x – 2y = 5

Adding we get,

5x = 15

⇒ x = 15/5 = 3

Substituting the values of x in (i)

2×3 + y = 5

⇒ 6 + y = 5

⇒ y = 5 – 6

= -1

∴ Co-ordinates of point of intersection are (3, -1)

∵ the line passes through (3, -1)

∴ -1 = m×3 – 3/7 **(y = mx + c)**

3m = -1 + 3/7 = -4/7

m = -4/(7×3)

= -4/21

∴ Equation of line y = -4/21 ×x – 3/7

⇒ 21y = -4x – 9

⇒ 4x + 21y + 9 = 0

**8. If point A is reflected in the y-axis, the co-ordinates of its image A _{1}, are (4, - 3). **

**(i) Find the co-ordinates of A. **

**(ii) Find the co-ordinates of A _{2}, A_{3} the images of the points A, A_{1}, Respectively under reflection in the line x = - 2 **

**Answer**

**(i) **∵ A is reflected in the y-axis and its image is A_{1} (4, -3)

Co-ordinates of A will be (-4, 3)

**(ii)** ∵ A_{2 }is the image of A (-4, -3) in the line

x = -2 which is parallel to y-axis

∴ AA_{2} is perpendicular to the line x = -2 and this line bisects AA_{2 }at M.

Let co-ordinates of A_{2} be (𝝰, 𝝱)

∴ 𝝰 = 0 and 𝝱 = -3

∴ Co-ordinates of A_{2} will be (0, -3)

Again x = -2 is perpendicular bisector of A_{1}, A_{3} intersecting it at M.

∴ M is mid point of A_{1}A_{3}.

∴ A_{3}M = MA_{1}

∴ Coordinates of A_{3} will be (-8, -3)

**9. If the lines x/3 + y/4 = 7 and 3x + ky = 11 are perpendicular to each other, find the value of k. **

**Answer**

Given equation of lines are x/3 + y/4 = 7

⇒ 4x + 3y = 84

⇒ 3y = -4x + 84

⇒ y = -4/3 ×x + 28 **…(i)**

And 3x + ky = 11

⇒ ky = -3x + 11

⇒ y = (-3/k)×x + 11/k **…(ii)**

Let slope of line (i) be m_{1} and of (ii) be m_{2}

∴ m_{1} = -4/3 and m_{2} = -(3/k)

∵ These lines are perpendicular to each other

∴ m_{1}m_{2} = - 1

⇒ -4/3 × (-3/k) = -1

⇒ 4/k = -1

⇒ - k = 4

⇒ k = - 4

**10. Write down the equation of a line parallel to x – 2y + 8 = 0 and passing through the point (1, 2). **

**Answer: **

The equation of the line is x – 2y + 8 = 0

⇒ 2y = x + 8

⇒ y = (1/2)×x + 4

∴ Slope of the line = 1/2

∴ Slope of the line parallel to the given line passing through (1, 2) = 1/2

∴ Equation of the lines will be,

y – y_{1} = m(x – x_{1})

⇒ y – 2 = 1/2 (x – 1)

⇒ 2y – 4 = x – 1

⇒ x – 2y – 1 + 4 = 0

⇒ x - 2y + 3 = 0

**11. Write down the equation of the line passing through (-3, 2) and perpendicular to the line 3y = 5 – x. **

**Answer**

Equation of the line is

3y = 5 – x

⇒ 3y = -x + 5

⇒ y = -(1/3)×x + 5/3

∴ Slope of the line = -1/3

And slope of the line perpendicular to it and passing through (-3, 2) will be = 3

(∵ m_{1}m_{2} = -1)

∴ Equation of the line will be

y – y_{1} = m(x – x_{1})

⇒ y – 2 = 3(x + 3)

⇒ y – 2 = 3x + 9

⇒ 3x – y + 9 + 2 = 0

⇒ 3x – y + 11 = 0

**12. Find the equation of the line perpendicular to the line joining the points A (1, 2) and B (6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2. **

**Answer**

Let slope of the line joining the points A (1, 2) and B (6, 7) be m_{1 }

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

= (7–2)/(6–1)

= 5/5

= 1

Let m_{2} be the slope of the line perpendicular to it then m_{1}×m_{2 }= -1

⇒ 1×m_{2} = -1

∴ m_{2} = -1

Let the point P (x, y) divides the line AB in the ratio of 3 : 2

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

= {3×6 + 2×1)/(3 + 2)

= (18+2)/5

= 20/5

= 4

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

= (3×7 + 2×2)/(3+2)

= (21+4)/5

= 25/5

= 5

∴ Co-ordinates of P will be (4, 5)

Now equation of the line passing through P and having slope –1

y – y_{1} = m(x – x_{1})

⇒ y – 5 = -1(x – 4)

⇒ y – 5 = -x + 4

⇒ x + y – 5 – 4 = 0

⇒ x + y – 9 = 0

**13. The points A (7, 3) and C (0, -4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.**

**Answer**

Slope of line AC (m_{1})

= (y_{2} – y_{1})/(x_{2 }– x_{1})

= (-4 – 3)/(0 – 7)

= -7/-7

= 1

∵ Diagonals of a rhombus bisect each other at right angles

∴ BD is perpendicular to AC

∴ Slope of BD = -1 **(∵ m _{1}m_{2 }= -1)**

And co-ordinates of O, the mid-point of AC will be {(7 + 0)/2, (3 – 4)/2} or (7/2, -1/2)

∴ Equation of BD will be

y – y_{1} = m(x – x_{1})

⇒ y + 1/2 = -1 (x – 7/2)

⇒ y + 1/2 = -x + 7/2

⇒ 2y + 1 = -2x + 7

⇒ 2x + 2y + 1 – 7 = 0

⇒ 2x + 2y – 6 = 0

⇒ x + y – 3 = 0 **(Dividing by 2)**

**14. A straight line passes through P (2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1, find : **

**(i) the co-ordinates of A and B.**

**(ii) the equation of the line AB **

**Answer**

**(i)** A lies on x-axis and B lies on y-axis

Let co-ordinates of A be (x, 0) and B be (0, y) and P (2, 1) divides BA in the ratio 3:1.

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

⇒ 2 = (3×x + 1×0)/(3+1)

⇒ 3x/4 = 2

⇒ 3x = 8

⇒ x = 8/3

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

⇒ 1 = (3×0 + 1×y)/(3+1)

⇒ y/4 = 1

⇒ y = 4

∴ Co-ordinates of A will be (8/3, 0) and of B will be (0, 4)

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

= (4 – 0)/(0 – 8/3)

= {4/(-8/3)}

= - 4 × 3/8

= -3/2

∴ Equation of AB will be

y – y_{1} = m(x – x_{1})

⇒ y – 1 = -3/2 (x – 2)

⇒ 2y - 2 = -3x + 6

⇒ 3x + 2y – 2 – 6 = 0

⇒ 3x + 2y – 8 = 0

**15. A straight line makes on the co-ordinates axes positive intercepts whose sum is 7. If the lines passes through the point (-3, 8), find its equation. **

**Answer**

Let the line make intercept a and b with the x-axis and y-axis respectively then the line passes through

A (a, 0) and B (0, b)

But a + b = 7

b = 7 – a

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

= (b – 0)/(0 – a)

= -b/a

∴ Equation of the line y – y_{1} = m(x – x_{1})

⇒ y – 0 = (-b/a).(x – a) **...(i)**

∵ the line passes through the point (-3, 8)

∴ 8 – 0 = -b/a(- 3 – a)

= - {(7 – a)/a}×(-3 – a)

⇒ 8a = (7 – a)(3 + a)

⇒ 8a = 21 + 7a – 3a + a^{2}

⇒ a^{2} + 8a – 7a + 3a – 21 = 0

⇒ a^{2} + 4a – 21 = 0

⇒ a^{2} + 7a – 3a – 21 = 0

⇒ a(a + 7) - 3(a + 7) = 0

⇒ (a + 7)(a – 3) = 0

Either a + 7 = 0, then a = -7, which is not possible as it is not positive

Or a – 3 = 0, then a = 3

and b = 7 – 3 = 4

∴ Equation of the line y – 0 = -(b/a)×(x – a)

⇒ y = -4/3(x – 3)

⇒ 3y = -4x + 12

⇒ 4x + 3y – 12 = 0

⇒ 4x + 3y = 12

**16. If the coordinates of the vertex A of a square ABCD are (3, - 2) and the equation of diagonal BD is 3x – 7y + 6 = 0, find the equation of the diagonal AC. Also find the co-ordinates of the center of the square.**

**Answer: **

Co-ordinates of A are (3, - 2)

Diagonals AC and BD of the square ABCD bisect each other at right angle at O.

∴ O is the mid-point of AC and BD equation

Diagonals AC and BD of the square ABCD bisect each other at right angle at O.

∴ O is mid-point of AC and BD

Equation of BD is 3x – 7y + 6 = 0

⇒ 7y = 3x + 6

⇒ y = (3/7)×x + 6/7

∴ Slope of BD = 3/7

And slope of AC = -7/3 **(∵ m _{1}m_{2} = -1)**

∴ Equation of AC will be

y – y_{1} = m(x – x_{1})

⇒ y + 2 = -7/3 (x – 3)

⇒ 3y + 6 = -7x + 21

⇒ 7x + 3y + 6 – 21 = 0

⇒ 7x + 3y – 15 = 0

Now we will find the co-ordinates of O, the points of intersection of AC and BD

We will solve the equations,

3x – 7y = - 6 **…(i)**

7x + 3y = 15 **…(ii)**

Multiplying (i) by 3 and (ii) by 7, we get,

9x – 21y = - 18 **…(iii)**

49x + 21y = 105 **…(iv)**

Adding we get,

58x = 87

⇒ x = 87/58 = 3/2

Substituting the value of x in (iii)

9(3/2) – 21y = - 18

⇒ 27/2 – 21y = - 18

21y = 27/2 + 18

= (27 + 36)/2

= 63/2

y = 63/(2×21)

= 3/2

∴ Co-ordinates of O will be (3/2, 3/2)

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