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

Here, we are providing the solutions for Chapter 12 Equation of Straight Line from ML Aggarwal Textbook for Class 10 ICSE Mathematics. Solutions of twelfth chapter has been provided in detail. This will help the students in understanding the chapter more clearly. Class 10 Chapter 12 Equation of Straight Line of ML Aggarwal Solutions for ICSE is one of the most important chapter for the board exams which is based on finding the mid-point of pairs of point, coordinates of the point and dividing the line in given ratio. We have also added chapter test and multiple choice questions.

### Exercise 12.1

1. Find the slope of a line whose inclination is

(i) 45°

(ii) 30°

(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

(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.

(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.

(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).

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

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

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.

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.

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).

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).

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.

∵ 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.

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)

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.

∴ 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) = (y2 – y1)/(x2 – x1)

= (-3+5)/(0 – 2)

= 2/-2

= - 1

∴ Equation of the line will be,

y – y1 = m(x – x1)

⇒ 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.

Equation of the line will be

y – y1 = m(x – x1)

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).

The equation of line whose slope is wand passes through a given point is y - y1 = m(x – x1)

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)

m = (y2 – y1)/(x2 – x1)

Given

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

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

(i) m = (4 + 2)/(3 – 0) = 6/3 = 2

(ii) m = (8 + 7)/(-1 – 3) = 15/-4

17. The coordinates of two points E and F are (0, 4) and (3, 7) respectively. Find:

(ii) The equation of EF

(iii) The coordinates of the point where the line EF intersects the x-axis.

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

∴ gradient (m) = (y2 – y1)/(x2 – x1)

= (7 – 4)/(3 – 0)

= 3/3

= 1

(ii) Equation of line EF,

y – y1 = m(x – x1)

⇒ 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.

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.

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

∴ Slope of the line (m) = (y2 – y1)/(x2 – x1) = (- 1 – 1)/(1 – 5)

= -2/-4

= 1/2

Equation of the line,

y – y1 = m(x – x1

⇒ 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.

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 – a2 – 2 + a

⇒ -6 = 3a – a2 – 2

⇒ a2 – 3a + 2 – 6 = 0

⇒ a2 – 3a – 4 = 0

⇒ a2 – 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).

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.

Given that

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

(i)

Co-ordinates of O = {(5+2)/2, (8–1 4)/2} = (3.5, 2)

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.

⇒ D is mid point of BC

∴ D is (7+1)/2, (8–10)/2}

i.e., (4, -1)

m = (y2 – y1)/(x2 – x1)

= (5+1)/(3–4) = 6/-1 = - 6

y – y1 = m(x – x1)

⇒ 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.

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) = (y2 – y1)/(x2 – x1)

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

= - 3/6

= - (1/2)

∴ Equation of the line will be y – y1 = m(x – x1)

⇒ 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.

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 = (y2 – y1)/(x2 – x1)

= (-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.

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 = (m1x2 + m2x1)/(m1 + m2)

= {2×3 + 3×(-2)}/(2+3)

= (6 – 6)/5

= 0/5

= 0

y = (m1y2 + m2y1)/(m1 + m2)

= {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 – y1 = m(x – x1)

⇒ 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.

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 = (y2 – y1)/(x2 – x1) = (- 11/2 – 4)/(0–1)

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

= 19/2

And equation of the line will be

y – y1 = m(x – x1)

⇒ 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.

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) = (y2 – y1) = (x2 – x1)

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

∴ Equation of the line will be

y – y1 = m(x – x1)

⇒ 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.

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.

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 = (y2 – y1)/(x2 – y1)

= (-6 – 0)/(0–4)

= -6/-4

= 3/2

(iii) Equation of AB will be y – y1 = m(x – x1)

⇒ 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.

The equations of sides of a rectangle whose equations are

x1 = - 1, x2 = 2, y1 = - 2, y2 = 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 = (y2 – y1)/(x2 – x1)

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

= 8/3

∴ Equation of AC will be

y – y1 = m(x – x1

= 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

5x + 7y = 3 …(i)

2x – 3y = 7 …(ii)

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

15x + 21y = 9

14x – 21y = 49

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 = (y2 – y1)/(x2 – x1)

= (0 + 1)/(0 – 2)

= 1/2

Equation of line will be

y – y1 = m(x – x1)

⇒ 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.

(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)

= (y2 – y1)/(x2 – x1) = (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.

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.

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

m1 = m2

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.

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 (m2) = -(a/3)

∵ Lines are parallel

∴ m1 = m2

⇒ 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.

In equation y = x - p

⇒ y = 2x – 2p

Slope (m1) = 2

In equation ax + 5 = 3y

⇒ y = a/3 ×x + 5/3

Slope (m2) = a/3

∵ Lines are parallel

∴ m1 = m2

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.

Gradient m1 of the line y = 3x + 7 is 3

2y + px = 3

⇒ y = (-px)/2 + 3/2

Gradient m2 of this line is – (p/2)

Since, the given lines are perpendicular to each other.

∴ m1 × m2 = - 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.

Given

In equation, 3x – 5y + 4 = 0

⇒ 5y = kx + 4

⇒ y = k/5 + 4/5

∴ Slope (m1) = k/5

And in equation, 4x – 2y + 5 = 0

⇒ 2y = 4x + 5

⇒ y = 2x + 5/2

Slope (m2) = 2

∵ Lines are perpendicular to each other

∴ m1m2 = - 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.

Given,

In the equation 3x + by + 5 = 0

by = -3x – 5

⇒ y = -3/b ×x – 5/b

Slope (m1) = -3/b

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

⇒ 5y = ax + 7

⇒ y = a/5 + 7/5

∴ Slope (m2) = a/5

∵ Lines are perpendicular to each other

∴ m1m2 = - 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).

Slope of the line passing through the points (- 2, 3) and (4, 1) = (y2 – y1)/(x2 – x1)

= (1 – 3)/(4 + 2)

= -2/6

= -1/3

Slope of line 3x = y + 1

∵ m1 × m2 = -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.

Gradient (m1) of the line passing through the points A (- 2, 3) and B (4, b)

= (b – 3)/(4 + 2)

= (b - 3)/6

Gradient (m2) of the line 2x – 4y = 5

Or y = x/2 – 5/2 is 1/2

Since, the lines are perpendicular to each other,

∴ m1 × m2 = - 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.

In the line 3x + y = 4 …(i)

⇒ y = -3x + 4

Slope (m1) = - 3

In the line x – ay + 7 = 0 …(ii)

⇒ ay = x + 7

⇒ y = (1/a)×x + 7/a

Slope (m2) = 1/a

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

⇒ 2y = -bx – 5

⇒ y = (-b/2) ×x – 5/2

∴ Slope (m3) = - b/2

∵ These are the consecutive three sides of a rectangle.

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

∴ m1m2 = - 1

⇒ -3 × 1/a = - 1

⇒ -3 = - a

⇒ a = 3

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

∴ m1 = m3

⇒ -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.

In the given line 2x – 3y – 7 = 0

⇒ 3y = 2x – 7

⇒ y = (2/3)×x – 7/3

Hence slope (m1) = 2/3

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

y – y1 = m(x – x1)

∵ 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.

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

⇒ 5y = -2x – 7

⇒ y = (-2/5)×x – 7/5

Here slope (m1) = -(2/5)

Let the slope of the line perpendicular to the given line = m2

∴ m1m2 = - 1

⇒ -(2/5)m2 = - 1

∴ m2 = -1 × -5/2

= 5/2

∵ It makes y-intercept –3 units

∴ The point where it passes = (0, -3)

∴ Equations of the new line,

y – y1 = m(x - x1)

⇒ 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.

In the given line 3x – 4y + 12 = 0

⇒ 4y = 3x + 12

⇒ y = (3/4)x + 3

Here slope (m1) = 3/4

Let the slope of the line perpendicular to the given line be = m2

∴ m1m2 = - 1

⇒ 3/4m2 = - 1

m2 = -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 – y1 = m(x – x1)

⇒ 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).

In the given line 3x – 2y = - 4

⇒ 2y = 3x + 4

⇒ y = 3/2×x + 2

Here slope (m1) = 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 – y1 = m(x – x1)

⇒ 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.

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

⇒ 5y = - 3x – 15

⇒ y = -3/5 ×(x – 3)

How slope (m1) = -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 – y1 = m(x – x1)

⇒ 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).

In the given line y = 3x – 5

Here slope (m1) = 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 – y1 = m(x – x1)

⇒ 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).

In the given line 3x + 8y = 12

⇒ 8y = - 3x + 12

⇒ y = -3/8 ×x + 12/8

Here slope (m1) = -3/8

Let the slope of the line perpendicular to the given line be = m2

∴ m1m2 = -1

⇒ -3/8 × m2 = -1

m2 = 8/3

∴ Equation of the line where slope is 8/3 and passes through the point (-1, -2) will be

y – y1 = m(x – x1)

⇒ 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.

(i) In the line 4x – 3y + 12 = 0 …(i)

3y = 4x + 12

⇒ y = 4/3 ×x + 4

Here slope (m1) = 4/3

Let the slope of the line perpendicular to the given line be = m2

∴ m1m2 = - 1

⇒ 4/3 ×m2 = - 1

⇒ m2 = -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 – y1 = m(x – x1)

⇒ 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).

The given line 2x + 5y – 7 = 0

5y = - 2x + 7

⇒ y = -2/5 ×x + 7/5

Here slope (m1) = -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 – y1 = m(x – x1)

⇒ 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).

In the given line 3x + 2y – 8 = 0

⇒ 2y = -3x + 8

⇒ y = -3/2 ×x + 4

Here slope (m1) = -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 = m2

∴ m1m2 = - 1

⇒ -3/2 m2 = - 1

⇒ m2 = 2/3

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

y – y1 = m(x – x1)

⇒ 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.

Slope (m1) of line by joining the points A (1, 3), B (3, - 1) = (y2 – y1)/(x2 – x1)

∴ m1 = (-1–3)/(3–1) = -4/2 = - 2

Slope (m2) of the line joining the points B

(3, -1) and C (-5, -5) = (y2–y1)/(x2–x1)

⇒ m2 = (-5+1)/(-5–3)

= (-4/-8)

= 1/2

∴ m1 × m2 = -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).

Slope of the line joining the points (0, - 2) and (4, 5) = (y2 – y1)/(x2 – x1)

= (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 – y1 = m(x – x1)

⇒ 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.

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

(i) Coordinates of centroid G = {(x1 + x2 + x3)/3, (y1 + y2 + y3)/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 = (y2 – y1)/(x2 – x1) = (-2 -3)/{(3–(-1)}

= -5/4

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

Let the equation of the line through G, be y – y1 = m(x – x1)

⇒ 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.

(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 = (y2 – y1)/(x2 – x1)

⇒ 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.

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).

Now slope of BC (m) = (y2 - y1)/(x2 – x1)

= (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 – y1 = m(x – x1)

⇒ 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).

The slope (m1) of the line joining the points (- 3, 2) and (9, 1)

= (y2 – y1)/(x2 – x1)

= (1 – 2)/(9 + 3)

= -1/12

Let slope of the line perpendicular to the line = m2

∴ m1m2 = -1

⇒ -1/12 × m2 = -1

⇒ m2 = -1 × (-12/1)

= 12

∴ Equation of the line passing through (0, -3) and of slope m2 = 12

y - y1 = m(x – x1)

⇒ 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.

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

Slope of the line BC (m1) = (y2 – y1)/(x2 – x1)

= (-1 + 9)/(-2 – 4)

= 8/(-6)

= -4/3

Let the slope of the altitude from A (10, 4) to BC = m2

∴ m1m2 = -1

⇒ -4/3 × m2 = -1

⇒ m2 = -1 ×(-3/4) = 3/4

∴ Equation of the line will be,

y – y1 = m(x – x1)

⇒ 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

(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)

(m) = (y2 – y1)/(x2 – x1)

= (4 + 4)/(1 – 2)

= 8/-1

= -8

Then equation of AD will be,

y – y1 = m(x – x1)

⇒ 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 (m1) = (y2 – y1)/(x2 - x1)

= (5 + 4)/(- 1 – 2)

= 9/-3

= -3

Let slope of BE = m2

But m1m2 = -1

⇒ - 3×m2 = -1

m2 = -1/-3 = 1/3

∴ Equation of BE will be,

y – y1 = m(x – x1)

⇒ 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).

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

m1 = (y2 – y1)/(x2 – x1)

= (-6 – 2)/(5 – 1)

= -8/4

= -2

Let m2 be the right bisector of the line

∴ m1m2 = -1

⇒ -2 × m2 = -1

m2 = (-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 – y1 = m(x – x1)

⇒ 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.

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

(i) ∴ Slope (m) = (y2 – y1)/(x2 – x1)

= {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 = (x1 + x2)/2, (y1 + y2)/2

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

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

∴ Slope of PQ = 1/2 (m1, m2 = -1)

∴ Equation of PQ = y – y1 = m(x – x1)

⇒ 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.

Slope of BD (m1) = (y2 – y1)/(x2 – x1)

= (8 – 3)/(6 – 1)

= 5/5

= 1

Diagonal AC is perpendicular bisector of diagonal BD

∴ Slope of AC = -1 (∵ m1m2 = -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 – y1 = m(x – x1)

⇒ 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.

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

Slope of AC (m1) = (y2 – y1)/(x2 – x1)

= (2 – 6)/(-1 – 3)

= -4/-4

= 1

∴ Slope of BD = - 1 (∵ m1m2 = -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 – y1 = m(x – x1)

⇒ 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.

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 (m1) = -1/2

∴ Slope of its parallel line = -1/2

And equation of the parallel line y – y1 = m(x – x1)

⇒ 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?

(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 = (m1x2 + m2x1)/(m1 + m2)

= {1×17 + 2×(-4)}/(1+2)

= (17 – 8)/3

= 9/3

= 3

y = (m1y2 + m2y1)/(m1 + m2)

= (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 m1 : m2

∴ x = (m1x2 + m2x1)/(m1+m2)

⇒ 0 = {m1×17 + m2×(-4)}/(m1+m2)

⇒ 17m1 – 4m2 = 0

⇒ 17m1 = 4m2

⇒ m1/m2 = 4/17

⇒ m1 : m2 = 4 : 17

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

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 (∵ m1m2 = -1)

∴ Equation of PP’

y – y1 = m(x – x1) = 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.

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 (∵m1m2 = -1)

And equation of line PQ

y – y1 = m(x – x1)

⇒ 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

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.

∵ OA = AB (sides of a square)

But OB2 = OA2 + AB2
⇒ p2 + q2 = 9 + (3 – p)2 + q2

⇒ p2 + q2 = 9 + 9 + p2 – 6p + q2

6p = 18

⇒ p = 18/6 = 3

Substituting the value of p in (i)

(3)2 + q2  - 6(3) = 0

⇒ 9 + q2 – 18 = 0

⇒ q2 – 9 = 0

⇒ q2 = 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

(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

(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

(c) -1

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

(y2 – y1)/(x2 – x1) = (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.

(a) 0

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

(y2 – y1)/(x2 – x1) = (-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

(d) not defined

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

(y2 – y2)/(x2 – x1) = (-4+2)/(3 – 3)

= -2/0

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

(a) 30°

(b) 60°

(c) 45°

(d) 0°

(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

(c) 1

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

m = (y2 – y1)/(x2 – x1)

⇒ 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

(b) -3/7

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

Slope of the line = (y2 – y1)/(x2 – x1)

= (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

(d) 5

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

= (y2 – y1)/(x2 – x1)

= (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

(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

(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

(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

(a) 3/2

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

∴ Product of their slopes = (m1×m2) = -1

Slope of 3x – 4y + 7 = 0

⇒ 4y = 3x + 7

⇒ y = 3/4 x + 7/4

Slope (m1) = 3/4

And slope of 2x + ky + 5 = 0

ky = -2x – 5

y = (-2/k) ×x – 5/k

∴ Slope (m2) = -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.

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.

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.

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.

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.

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 m1 : m2

∴ x = (m1x2 + m2x1)/(m1 + m2)

⇒ 3 = (m1p + m2×1)/(m1 + m2)

3m1 + 3m2 = m1p + m2

⇒ 3m1 – m1p = m2 – 3m2

⇒ m1(3 – p) = -2 m2

⇒ m1/m2 = - 2/(3 – p) …(i)

and –2 = {m1(-5) + m2×4}/(m1 + m2)

⇒ -2m1 – 2m2 = - 5m1 + 4m2

⇒ -2m1 + 5m1 = 4 m2 + 2m2

⇒ 3m1 = 6 m2

⇒ m1/m2 = 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).

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

= (y2 – y1)/(x2 – x1) = (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.

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

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 A1, are (4, - 3).

(i) Find the co-ordinates of A.

(ii) Find the co-ordinates of A2, A3 the images of the points A, A1, Respectively under reflection in the line x = - 2

(i) ∵ A is reflected in the y-axis and its image is A1 (4, -3)

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

(ii) ∵ A2 is the image of A (-4, -3) in the line

x = -2 which is parallel to y-axis

∴ AA2 is perpendicular to the line x = -2 and this line bisects AA2 at M.

Let co-ordinates of A2 be (𝝰, 𝝱)

∴ 𝝰 = 0 and 𝝱 = -3

∴ Co-ordinates of A2 will be (0, -3)

Again x = -2 is perpendicular bisector of A1, A3 intersecting it at M.

∴ M is mid point of A1A3.

∴ A3M = MA1

∴ Coordinates of A3 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.

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 m1 and of (ii) be m2

∴ m1 = -4/3 and m2 = -(3/k)

∵ These lines are perpendicular to each other

∴ m1m2 = - 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).

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 – y1 = m(x – x1)

⇒ 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.

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

(∵ m1m2 = -1)

∴ Equation of the line will be

y – y1 = m(x – x1)

⇒ 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.

Let slope of the line joining the points A (1, 2) and B (6, 7) be m1

∴ m1 = (y2 – y1)/(x2 – x1)

= (7–2)/(6–1)

= 5/5

= 1

Let m2 be the slope of the line perpendicular to it then m1×m2 = -1

⇒ 1×m2 = -1

∴ m2 = -1

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

∴ x = (m1x2 + m2x1)/(m1 + m2)

= {3×6 + 2×1)/(3 + 2)

= (18+2)/5

= 20/5

= 4

And y = (m1y2 + m2y1)/(m1 + m2)

= (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 – y1 = m(x – x1)

⇒ 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.

Slope of line AC (m1)

= (y2 – y1)/(x2 – x1)

= (-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 (∵ m1m2 = -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 – y1 = m(x – x1)

⇒ 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

(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 = (m1x2 + m2x1)/(m1 + m2)

⇒ 2 = (3×x + 1×0)/(3+1)

⇒ 3x/4 = 2

⇒ 3x = 8

⇒ x = 8/3

And y = (m1y2 + m2y1)/(m1 + m2)

⇒ 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 = (y2 – y1)/(x2 – x1)

= (4 – 0)/(0 – 8/3)

= {4/(-8/3)}

= - 4 × 3/8

= -3/2

∴ Equation of AB will be

y – y1 = m(x – x1)

⇒ 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.

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 = (y2 – y1)/(x2 – x1)

= (b – 0)/(0 – a)

= -b/a

∴ Equation of the line y – y1 = m(x – x1)

⇒ 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 + a2

⇒ a2 + 8a – 7a + 3a – 21 = 0

⇒ a2 + 4a – 21 = 0

⇒ a2 + 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.

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 (∵ m1m2 = -1)

∴ Equation of AC will be

y – y1 = m(x – x1)

⇒ 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)

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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