## Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 10 Straight Lines

**Short Answer** **Type Questions **

**(Score 3)**

Question 1.

a. Distance of the point (x, y) from the origin

a.x

b.y

c.x + y

d.

b. Find the angle between the x-axis and the line joining the points (3,-1) and (4,-2).

Answer:

a.

Find the angle between the x-axis and the line joining the points (3,-1) and (4,-2).

b. Given (x_{1}, y_{1}) = (3,-1) and (x_{2}, y_{2}) = (4,-2)

tanθ=-l =-tan 45= tan (180-45)

= tan 135

θ=135°

Question 2.

i. Find the distance between the pair of parallel lines y = mx + c and y = mx + d.

ii. Find the distance between the pair of lines 15x + 8y – 34 = 0 and 30x + 16y + 62 = 0.

Answer:

i. Lines are mx – y + c = 0 and mx-y+d=0

.

Question 3.

a. Find the distance between the lines 4x + 3y = 11 and 8x + 6y = 15.

b. (0,0) is a vertex of a square and 5x – 12y + 26=0 is one of the equation of its sides. Find the area of the square

Answer:

a. Lines are 4x+3y=11 and 8x+6y=15

4x + 3y= 15/2

Distance between parallel lines

b. Distance of the point P(0,0) from the line 5x -12y + 26 =0 is given by

Length of one side of square = 2 units Area of square = 2^{2} = 4 sq.units

Question 4.

a. Find the slope of a line perpendicular to the line which passes through (0, 8) and (-5,2)

a.

b.

c.

d.

b. Find the distance between the parallel lines 5x – 12y – 14 = 0 and 5x – 12y + 12 = 0.

Answer:

a.

slope of line which is perpendicular to line having slope m is given by

Question 5.

a. Identify the figure in which the line has a positive slope.

b. Find the x and y intersects of the line 3x + 4y – 12 = 0.

Answer:

a. Figure II

b. 3x + 4y=12 x – intersept = 4 y – intersept = 3

Question 6.

Coordinates of the foot of perpendicular from (a, 0) on the line y = mx + are m

a. (a. -a/m)

b. (a/m, 0)

c. (0, a/m)

d. (0,0)

Answer:

Question 7.

a Lines 2x – 3y = 5 and 6x – 9y – 7 = 0 are

a. perpendicular

b. parallel

c. intersecting but not perpendicular

d. coincident

b. A straight line perpendicular to the line 2x + y = 3 is passing through (1,1). Find its y-intercept.

Answer:

Question 8.

Find the equation of the perpendicular bisector of the line segment joining A (-2, 3) and B (6, -5).

Answer:

Equation of the line with slope -1 and passing through the point (2,-1) is y + 1 = 1 (x-2) y+1 = x-2,;

x-y – 3 = 0, x-y = 3

Question 9.

a. Consider the points A (6, 2) B (3, -1) and C(-2,4).

i. Find AB,BC and AC.

ii. Show that Δ ABC is right angled.

Answer:

Question 10.

a. Find the x and y intercepts of the line 6x + 3y – 5 = 0.

b. Angle between two lines is π/2.and the slope of one line is 1/2. Find the slope of

Answer:

Question 11.

a. Write the equation of Y-axis.

b. Find the distance between the lines 8x+ 15y -5 = 0 and 8x + 15y +12 = 0.

Answer:

a. x=0

**Short Answer** **Type Questions **

**(Score 4)**

Question 1.

Consider the lines 2x + y = 1 and 4x – y = 5.

i. Find the slopes of these lines.

ii. Find the tangent of angle between them.

iii. Find the equation of the line through the intersection of the given lines and passing through (5, -3).

Answer:

i. Slope of 2x + y = 1 is -2

Slope of 4x-y = 5 is 4

ii.

.

iii. Required family of lines is

2x + y -1 + λ (4x – y – 5) = 0—-

(1) passes through (5, -3)

2×5-3-1+λ(4×5 + 3-5) = 0

i…e, λ=-1/3

Put in (1)

2x + y-1 + (4x-y-5) = 0

6x + 3y-3-4x + y + 5= 0

2x + 4y + 2 = 0

x + 2y + 1 =0

Question 2.

Consider the figure given below:

1. Find coordinates of A, B and C.

2. Find the centroid of Δ ABC.

3. Find the area ofΔ ABC.

Answer:

5x-y-11=0………………………. (1)

x-2y+5 = 0……………………. (2)

4x+y-7 = 0……………………. (3)

Solving (1) and (2) A is (3,4)

Solving (1) and (3) B is (2, -1)

Solving (2) and (3) C is (1,3)

Question 3.

A (-7, -3), B (3, -5), C (5,10) are three vertices of the parallelogram of which AB and AC are the sides.

i. Find the fourth vertex.

ii. Calculate the length of the diagonals.

Answer:

Question 4.

Consider the line 4x – 3y +12 = 0

a. Find the equation of the line passing through the point(l,2) and parallel to the given line.

b. Find the distance between these two parallel lines.

c. Which among the following lines is perpendicular to the line 4x – 3y +12 = 0?

i. 2x + 3y – 8 = 0

ii. 4x – 3y + 5 = 0

iii. x + y = 7

iv. 3x + 4y + 9 = 0

Answer:

a. 4x-3y + k=0

Passing through (1,2)

Required equation 4x – 3y + 2 = 0

b.

c. 3x + 4y+9 = 0

Question 5.

The slopes of the lines which makes an angle 45° with the line 3x – y = -5 are

a. 1,-1

b. 1/2,-1

c. -2,1/2

d. 1,1

Answer:

Question 6.

Consider the lines x + y + 1 = 0 and x-y+1=0

i. Find the intercepts of the line x + y + 1 = 0 with the axes.

ii. Find the angle between the lines.

iii. Find the equation of the line passing through the intersection of the given lines which has

y-intercept 4.

Answer:

i. x + y+ 1 =0, -x-y= 1, intercepts are -1 and -1

ii. Slope of x + y+ 1 = 0 is -1

Slope of x-y+1=0 is 1

Product of the slopes = -1

∴ lines are perpendicular ,angle θ=90°

iii. Family of lines is

x + y+1+A(x + y+1) = 0……

(1) passes through (0, 4) i.e,4+1+A(-4+1) = 0

5-3λ = 0, λ =

Put λ = in (1) and proceed as in the previous question.

Equation of the line is 4x – y + 4 = 0

Question 7.

Let A (2, 3), B (4, -1) and C (-1, 2) be the vertices of Δ ABC.

i. Find the equation of BC.

ii. Find the foot of the altitude from A.

Answer:

Question 8.

By using the concept of equation of the line, prove that the three points (3,0), (-2, -2) and are collinear.

Answer:

Let A(3,0), B (-2, -2) and C (8,2) be the given mints.

4x+8= 10y+20

4x- 10y-12 = 0

2x – 5y – 6 = 0

∴ AB and BC represents the same line A, B and C are collinear.

Question 9.

i. Find the equation of the line passing through the point of intersection of

x – 7y + 5 = 0 and 3x + y – 7 =0

ii. If the line is parallel to x-axis, then find the equation of the line.

Answer:

i. The equation of the line passing through the point of intersection of the lines is of the form

Question 10.

Show that the equations of the lines passing through the intersection of the lines

x – 3y + 1=0, 2x + 5y – 9 = 0 and at a distance 2 units from origin are x = 2,3x + 4y = 10

Answer:

Equation to the required line is

x – 3y +1 + λ(2x + 5y – 9) = 0

i.e,(l +2 λ)x + (5 λ-3)y + 1 -9 λ = 0………….. (1)

∴ perpendicular distance of (1) from the

Question 11.

a. Find the slope of the line = 1

b. If the lines joining the points (0,0), (1,1) and (2,2), (4, y) are perpendicular, find y.

Answer:

Question 12.

a. Find the angle between the x-axis and the line joining (2, -1) and (4, -3).

b. Convert the equation of the line 2x-3y + 6 = 0 into intercept form.

Answer:

Question 13.

a. Find the distance between the pair of lines 4x – 3y – 9 = 0 and 8x – 6y – 21 = 0

b. Find the distance of the point (3, -3) from the line 3x – 4y – 26 = 0.

Answer:

Question 14.

i. Find the slope of the line through the points (1,2) and (4,2).

ii. What is the centroid of the triangle whose vertices are (x_{1}, y_{1}) (x_{2}, y_{2}) and (x_{3}, y_{3})?

iii. Find the relation between x and y such that (x, y) is equidistant from (6, -1) and (2,3)

Answer:

Question 15.

i. Point of concurrence of the medians of a triangle is called

ii. Show that the points (-1, -1) (2, 3) and (8,11) are collinear.

Answer:

a. Lines are 4x – 3y – 9 = 0 and

ii. Let A(-1,-1),B(2,3), C(8,11) be the points

Question 16.

i. What is the equation of the line perpendicular to the Y-axis and intersecting it at a point 3 units below the origin.

ii. Find the distance between the parallel lines 4x – 3y – 9 = 0 and 4x – 3y – 24 = 0

Answer:

i. Given point is (0, -3)

Required line is y = -3

i.e, y + 3 = 0

.

**Long Answer** **Type Questions **

**(Score 6)**

Question 1.

Find the points on the line y = 3x + 5 whose distances from the line 12x – 5y+22 = 0 are 6.

Answer:

Given lines are y = 3x + 5…….. (1) and

12x-5y + 22 = 0…………….. (2)

Let P(x_{1} , y_{1} ) be a point on (1) which is at a distance 6 unit from (2)

Since P is a point on (1), we get

y_{1} = 3x_{1}+5……………….. (3)

Perpendicular distance of P from (2) = 6

Question 2.

In the figure let P’ be the image of P (2,1) with respect to the line x+y – 5 = 0

i. Find the slope of PP’.

ii. Find the coordinates of the foot of perpendicular from P.

iii. Find the image of P i.e, P’

Answer:

i. Slope of x+y-5=0 is

∴ Slope of PP’= =1

ii. Equation of PP’ is y -1 = 1 (x – 2)

y -1 = x – 2 i.e, x – y -1 = 0

Foot of perpendicular is the point of intersection of x + y- 5 = 0 and x – y – 1 = 0. Solving we get

x = 3,y = 2

∴ M (3,2) is the foot of perpendicular.

iii. Let P’ be the point (x, y). M is the midpoint of PP’

Question 3.

Consider two lines whose intercepts on the axes are respectively p, – q and q, -p.

i. Find the equation of the lines.

ii. Find the tangent of the angle between the lines.

iii. Find the point of intersection of the lines.

Answer:

Equation of the line whose intercepts are p and -q is

i.e, -qx + py + pq = 0——- (1)

similarly equation of the other line is

-px + qy + pq = 0 …………………(2)

ii. Slope of the line (1) is –

Slope of the line (2) is

Question 4.

Consider the lines 5x – 3y = 1,2x + 3y = 23, 42x + 21y = 257

i. Prove that the lines are concurrent.

ii. Find the point of concurrency.

Answer:

i. 5x-3y=1

⇒ 5x-3y-1=0……………… (1)

2x + 3y=23

⇒ 2x + 3y-23=0…………….. (2)

42x+21y=257

⇒ 42x + 21y-257 = 0……….. (3)

Solving (2) and (3), we get

Question 5.

Vertices of A ABC are A(-1, 3), B (2, -1) and C (0,0) AD and BE are the altitudes.

i. Find the equations of BC and AC.

ii. Find the coordinates of the orthocentre.

Answer:

By point slope form we have y- y_{1} = m (x – x_{1})

Equation of AD is y – 3 = 2 (x – (-1))

Orthocentre is the point of intersection of the altitudes AD and BE,

Equation of AD is 2x-y+5 = 0……………… (1)

Equation of BE is x -3 y – 5 = 0………… (2)

Point of interaction is obtained by solving (1) and (2)

Question 6.

Consider the points A(0, 0 ), b(4, 2) and C(8,0).

a. Find the mid-point of AB.

b. Find the equation of the perpendicular bisector of AB.

c. Find the equation of the circumcircle ( circle passing through the points A, B and C) of triangle ABC.

Answer:

a.(2,1)

Question 7.

a. Consider the straight line passing through A (-2,6) and B (4,8).Find the slope of the line AB.

b. Find the equation of the line which cut off intercepts on the axes, whose sum and product are 1 and -6 respectively,

c. What are the points on the axis of y whose perpendicular distance from the line 4x- 3y -12 = 0 is 3.

Answer:

Question 8.

Consider two points A (-2,-3) and B (1,6).

i. Find the equation of the line passing through A and B.

ii. Find the equation of the line passing through (2,1) and perpendicular to .

iii. Find the foot of the perpendicular.

Answer:

9x+ 18 = 3y + 9

9x-3y + 9 = 0

3x-y + 3 = 0

ii. Any line perpendicular to AB is x + 3y + k = 0 this line passes through (2,1) i.e,2 + 3 x 1 +k=0 5 + k = 0,k=-5

∴ Required line is x + 3y-5 = 0

iii. Foot of the perpendicular is the point of intersection of 3x-y + 3= 0 and x + 3y-5=0

Solving x =9/5, y = -2/5

Foot of perpendicular is

Question 9.

i. The coordinate of the point which divides (1, -3) and (-3,9) internally in the ratio 1: 3 is

a. (1,0)

b. (0,1)

c. (0,0)

d. (1,1)

ii. For what value of x the points(x, -1) (2,1) (4,5) are collinear

iii.In the figure given equation of is

Answer:

Question 10.

a. Find the equation of the line passing through the point (-1,2) and having sloped

b. The line through (4, 3) and (-6, 0) intersects the line 5x + y = 0. Find the angle of intersection.

c. Find the equation of the line that has y intercept -3 and is perpendicular to the line 3x + 5y = 4.

Answer:

Any line perpendicular to 3x + 5y-4 = 0 is

5x-3y + k=0………………….. (1)

y intercept = -3

i..e, line passes through (0,-3)

substituting in (1)

5×0-3x-3+k = 0

0 + 9 + k = 0, k = -9

Equation of the required line is 5x-3y- 9 = 0

Question 11.

i. Find the slopes of the lines 3x + 4y + 13 and 12x – 5y + 32 = 0

ii. Find the equation of the line through the intersection of these lines and passing through origin

Answer:

Slope of 3x+4y +13 = 0 is -3/4

Slope of 12x – 5y+32 = 0 is 12/5

ii. Any line through the intersection of 3x+4y+ 13 = 0 and 12x-5y+32 = 0

is 3x+4y+13 + λ (12x-5y+32)=0………….. (1)

This line passes through origin (0,0)

13 + λ .32 = 0,

32λ = -13, λ =

Sub this in (1)

3x + 4y+13+ (12x-5y+32) = 0

32(3x+4y+13)-13 (12x-5y+32) = 0

-60x+193y=0,

60x- 193y=0

**NCERT Questions and Answers**

Question 1.

Find the slope of the line making inclination of 60° with the positive direction of x-axis.

Answer:

Here inclination of the line α = 60°. Therefore, slope of the line is m = tan 60° =

Question 2.

If the angle between two lines is and and slope of one the lines is .find the slope of the other line.

Answer:

Question 3.

Find the equation of the line through (- 2,3) with slope – 4.

Answer:

Here m=- 4 and given point (x_{0} -y_{0}) is (- 2,3).

By slope-intercept form formula of the given line is

y-3=-4(x + 2) or 4x + y+5=0,which is the required equation.

Question 4.

A person standing at the junction of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 =0 in the least time. Find the equation of the path that he should follow.

Answer:

M is obtained by solving 2x – 3y + 4 = 0 and 3x+4y-5 = 0

To reach the path 6x – 7y + 8 = 0 in the least time, he should follow the path MS slope of

Question 5.

Write the equation of the line through the points (1, -1) and (3,5).

Answer:

Here x_{1}= 1, y_{1}=-1, x_{2}=3 and y_{2 }=5.using two-point form (2) above for the equation of the line, we have

-3x+y+4=0, Which is the requied equation.

Question 6.

Find the equation of the line through (-2,1) and making 45^{0} with the positive direction of x-axis in

i. Symmetrical form.

ii. Point slope form.

Answer:

ii. m=tan 45 = 1

Equation of line is y – y_{1} = m (x – x1)

y -1 = 1 (x – -2)

y-1 =x + 2

x-y+3 = 0

Question 7.

Line through the points (-2,6) and (4,8) is perpendicular to the line through the points (8,12) and

(x, 24). Find the value of x.

Answer:

Slope of the line through the points(-2,6)

Question 8.

Find the value of x for which the points (x,-l), (2,1) and (4,5) are collinear.

Answer:

Given points, A (x, -1), B (2,1) and C (4,5) are collinear.

Question 9.

A ray of light coming from the point (1,2) is reflected at a point A on the X-axis and then passes through the point (5, 3). Find the coordinates of the point A.

Answer:

Let the incident ray strike X-axis at the point A, whose coordinates be (x, 0).

The slope of the reflected ray is given by tan

Question 10.

If three points (h, 0), (a, b) and (0, k) lie on a line. Show that

Answer:

Given points are A (h, 0), B (a, b) and C (0, k). If A, B and C are collinear, the slope of

AB = slope of BC = slope of CA.