Miscellaneous Exercise Page: 233

1: Find the value of k for which the line (k – 3)x-(4 - k 2 )y + k 2 - 7k + 6 = 0 is (a) Parallel to x-axis,(b) Parallel to y-axis,(c) Passing through the origin.

Solution:

The given equation of line is (k – 3)x-(4 - 2k )y + 2k - 7k + 6 = 0 …(1) (a) If the given line is parallel to the x-axis , thenSlope of the given line = Slope of the x-axisThe given line can be written as

2 2(4-k )y=(k-3)x+ k 6 07k

2

2 2(k-3) k 7 6(4-k ) (4-k )

kxy , which is of the form y = mx + c.

2

2

(k 3)Slope of the given line =(4 )

Slope of the x-axis = 0(k 3) 0(4 k )

03

k

k

k 3

Thus, the given line is parallel to x-axis , then the value of k is 3. (b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined.

The slope of the given line is 2(k 3)(4 )k

Now, 2(k 3)(4 )k

is undefined at 2k = 4

2k = 4

NCERT Solution For Class 11 Maths Chapter 10 Straight Lines

Miscellaneous Exercise Page: 233

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k = 2 Thus, if the given line is parallel to the y-axis, then the value of k is 2. (c) If the given line is passing through the origin, then point (0, 0) satisfies the given equationof line. 2 23 0 (4-k ) 0 k 7 6 0k k

2

2

k 7 6 0k 6 6 0

6 1 0

kk k

k k

k = 1 or 6 Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.

2: Find the values of and p, if the equation x cos + y sin = p is the normal form of the line

3 2 0x y

3 2 0x y

Solution:

The equation of the given line is This equation can be reduced as

3 2 0x y

3 2x y

On dividing both sides by 2 23 1 2 , we obtain

3 1 22 2 2

3 1 1 ... 12 2

x y

x y

On comparing equation (1) to x cos + y sin = p, we obtain

cos = - 32

, sin = 12

, and p =1

Since the value of sin and cos are negative, 76 6

Thus, the respective values of and p are 76 and 1.

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

Solution:

Let the intercepts cut by the given lines on the axes be a and b.




It is given that a + b = 1 … (1) ab = -6 … (2) On solving equations (1) and (2) , we obtain a = 3 and b = -2 or a = -2 and b = 3 It is known that the equation of the line whose intercepts on the axes are a and b is

1x ya b or bx + ay – ab = 0

Case I: a = 3 and b = -2 In case, the equation of the line is -2x + 3y + 6 = 0, i.e., 2x – 3 y = 6. Case II: a = - 2 and b = 3 In this case, the equation of the line is 3x – 2y + 6 = 0, i.e., -3x + 2y = 6. Thus, the required equation of the lines are 2x – 3 y = 6 and -3x + 2y = 6.

4:

What are the points on the y-axis whose distance from line 13 4x y is 4 units.

Solution:

Let (0, b) be the point on y-axis whose distance from line 13 4x y is 4 units.

The given line can be written as 4x + 3y – 12 = 0 … (1) On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, C = -12. It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point 1 1,x yis given by

1 1

2 2

Ax By Cd

A B

Therefore, if (0, b) is the point on the y-axis whose distance from line 13 4x y is 4 units,

then:

2 2

4 0 3 124

4 33 12

45

20 3 12

20 3 12

20 3 12 20 3 123 20 12 3 20 12

32 83 3

b

b

b

b

b or bb or b

b orb




Thus, the required points are 320,3

and 80,3

5: Find the perpendicular distance from the origin to the line joining the points (cos θ,sin) θand (cosφ,sin) φ

Solution:

The equation of the line joining the points is given by

cos ,sin cos ,sin

sin sinsin coscos cos

cos cos sin cos cos sin sin cos sin sin

sin sin cos cos cos sin cos sin sin cos sin cos 0

sin sin cos cos sin( ) 0

and

y x

y x

x y

x yAx By C

0, sin sin , cos cos , sin( )whereA B andC

It is known that the perpendicular distance (d) of a line Ax +By +C =0 from a point 1 1,x y is given by

1 1

2 2

Ax By Cd

A B

Therefore, the perpendicular distance (d) of the given line from point 1 1,x y = (0, 0) is




2 2

2 2 2 2

2 2 2 2

2

sin sin 0 cos cos 0 sin

sin sin cos cos

sin

sin sin 2sin sin cos cos 2cos cos

sin

sin cos sin cos 2 sin sin cos cos

sin

1 1 2 cos

sin

2 1 cos

sin

2 2sin

d

2

sin

2sin2

6: Find the equation of the line parallel to y-axis and draw through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y =0.

Solution:

The equation of any line parallel to the y-axis is of the form x = a … (1) The two given lines are x – 7y + 5 = 0 … (2) 3x + y = 0 … (3)

On solving equation (2) and (3), we obtain x = - 522

and y = 1522

Therefore, 5 15,22 22

is the point of intersection of lines (2) and (3).

Since line x = a passes through point 5 15,22 22

, a = - 522

Thus, the required equation of the line is x = - 522

7:




Find the equation of a line drawn perpendicular to the line 4 6x y = 1 through the point, where

it meets the y-axis.

Solution:

The equation of the given line is 4 6x y = 1

This equation can also be written as 3x + 2y – 12 = 0

y= 32 x + 6, which is of the form y = mx + c

Slope of the given line = 32

Slope of line perpendicular to the given line = 1 23 32

Let the given line intersect the y-axis at (0, y).

On Substituting X with 0 in the equation of the given line, we obtain 1 66y y

The given line intersects the y-axis at (0, 6).

The equation of the line that has a slope of 23

and passes through point (0, 6) is

26 ( 0)3

3 18 22 3 18 0

y x

y xx y

Thus, the required equation of the line is 2 3 18 0x y

8: Find the area of the triangle formed by the line y – x = 0, x + y = 0 and x – k = 0.

Solution:

The equation of the given lines are y – x = 0 … (1) x + y = 0 … (2) x – k = 0 … (3) The point of intersection of lines (1) and (2) is given by x = 0 and y = 0 The point of intersection of lines (2) and (3) is given by x = k and y = -k The point of intersection of lines (3) and (1) is given by x =k and y = k Thus, the vertices of the triangle formed by the three given lines are (0, 0) ,(k, -k), and (k, k). We know that the area of a triangle whose vertices are 1 1,x y , 2 2 3 3, , ,x y and x y is




1 2 3 2 3 1 3 1 212

x y y x y y x y y

Therefore, area of the triangle formed by the three given lines

2 2

2

2

1 0 0 02121 22

k k k k k k square units

k k square units

k square units

k square units

9: Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.

Solution:

The equation of the given lines are 3x + y – 2 = 0 … (1) px + 2y – 3 = 0 … (2) 2x – y – 3 = 0 … (3) On solving equations (1) and (3), we obtain x = 1 and y = -1 Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2). p (1) + 2(-1) -3 = 0 p – 2 – 3 =0 p = 5 Thus, the required value of p is 5.

10: If three lines whose equations are 1 1 2 2 2 2, ,y m x c y m x c and y m x c are concurrent, then show that 1 2 3 2 3 1 3 1 2 0m c c m c c m c c

Solution:

The equation of the given lines are 1 1y m x c … (1)

2 2y m x c … (2)

2 3y m x c … (3) On subtracting equation (1) from (2) we obtain




2 1 2 1

1 2 2 1

2 1

1 2

0 m m x c c

m m x c cc cx

m m

On substituting this value of x in (1), we obtain

2 11 1

1 2

1 2 1 11

1 2

1 2 1 1 1 1 2 1

1 2

1 2 2 1

1 2

c cy m cm m

m c m cy cm m

m c m c m c m cym m

m c m cym m

2 1 1 2 2 1

1 2 1 2

,c c m c m cm m m m

is the point of intersection of line (1) and (2).

It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3).

1 2 2 1 2 13 3

1 2 1 2

3 2 3 1 3 1 3 21 2 2 1

1 2 1 2

1 2 2 1 3 2 3 1 3 1 3 2

1 2 3 2 3 1 3 1 2

1 2 3 2 3 1 3 1 2

00

Hence, 0

m c m c c cm cm m m m

m c m c c m c mm c m cm m m m

m c m c m c m c c m c mm c c m c c m c c

m c c m c c m c c

4511:

Find the equation of the line through the points (3, 2) which make an angle of with the line x – 2y = 3.

1m . Solution:

Let the slope of the required line be

The given line can be written as y = 1 32 2

x , which is of the form y= mx + c

Slope of the given line = 212

m

It is given that the angle between the required line and line x – 2y = 3 is 45 . We know that if is the acute angle between lines 1 2l and l with slopes 1 2m and m respectively, then




tan = 2 1

1 21m m

m m

tan 45 = 2 1

1 21m m

m m

1

1

1

1

1

1

1

1

1 1

1 1

1211

21 2

21 22

1 212

1 212

1 2 1 21 12 2

m

m

m

m

mm

mm

m morm m

1 1 1 1

1 1

1

1

2 1 2 2 1 21 33

Case I: 3The equation of the line passing through (3,2) and having a slope of 3 is:y-2=3(x-3)y-2=3x-93x-y=7

1Case II: 3

The equation of the line passing throu

m m or m m

m or m

m

m

1gh (3,2) and having a slope of :3

1y-2= ( 3)3

3 6 33 9

Thus, the equations of the line are 3x-y =7 and 3 9.

is

x

y xx y

x y

12: Find the equation of the line passing through the point of intersection of the line 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.




Solution:

Let the equation of the line having equal intercepts on the axes be

1

Or x+y=a ...(1)

x ya a

On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we obtain 1 5x=13 13

and y

1 5,

13 13

is the point of the intersection of the two given lines.

Since equation (1) passes through point 1 5,13 13

,

1 513 13

613

a

a

Equation (1) becomes x + y = 613

, i.e., 13x + 13y = 6.

Thus, the required equation of the line 13x+13y=16.

13: Show that the equation of the line passing through the origin and making an angle with the line

y = mx + c , is yx

= tan1 tanm

m

Solution 13:

Let the equation of the line passing through the origin be y = 1m x . If this line makes an angle of with line y = mx + c, then angle is given by

1

1

tan1

1

tan1

tan1

1

m mm m

y mx

y mxy mx

y mx

yymm

x xor yy mmxx

y mx

y mx

tan

tan

tan

1




Case I:

tan1

tan tan

tan (1 tan )

tan1 tan

Case II: tan1

tan1

tan tan

1 tan tan

tan1 tan

Therefore,

y mx

y mxy ym mx x

ym mx

y mx m

x my

y mx

y mx

y mx

y ym mx x

y m mxy mx m

tanthe required line is given by 1 tan

y mx m

14: In what ratio, the line joining (-1, 1) and (5, 7) is divisible by the line x + y = 4?

Solution:

The equation of the line joining the points (-1, 1) and (5, 7) is given b y

7 11 15 161 162 0 ... 1

y x

y x

x y

The equation of the given line is X + y – 4 = 0 …(2) The points of intersection of line (1) and (2) is given by x = 1 and y = 3




Let point (1, 3) divide the line segment joining (-1, 1) and (5, 7) in the ratio 1 : k. Accordingly, by section formula,

1 1(5) 1 1 71,3 ,

1 1

5 71,3 ,1 1

5 71, 31 1

5 11

5 12 4

2

k kk k

k kk k

k kk k

kk

k kk

k

Thus, the line joining the points (-1, 1) and (5, 7) id divided by line x + y = 4 in the ratio 1:2.

15: Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.

Solution:

The given lines are 2x – y = 0 … (1) 4x + 7y + 5 = 0 … (2) A (1, 2) is a point on line (1). Let B be the point intersection of line (1) and (2).

On solving equations (1) and (2), we obtain 5 518 9

x and y

Coordinates of point B are 5 5,18 9

.

By using distance formula, the distance between points A and B can be obtained as




2 2

2 2

2 2

2 2 2

2

5 51 218 9

23 2318 9

23 232 9 9

23 1 239 2 9

23 1 19 4

23 59 423 59 223 5

18

AB units

units

units

units

units

units

units

units

Thus, the required distance is 23 518

units .

16: Find the direction in which a straight line must be drawn through the points (-1, 2) so that its point of intersection with line x + y = 4 may be at a distance of 3 units from this point.

Solution:

Let y = mx + c be the line through point (-1, 2). Accordingly, 2 = m(-1) + c.

22

2 ...(1)

m cc m

y mx m

The given line is x + y = 4 … (2) On solving equation (1) and (2), we obtain

2 5 21 1m mx and y

m m

2 5 2,1 1m m

m m

is the point of intersection of line (1) and (2).

Since this point is at a distance of 3 units from points (-1, 2), accordingly to distance formula,




2 2

2 22

2

2 2

2

2

2 2

2 5 21 2 31 1

2 1 5 2 2 2 31 1

9 9 91 1

1 11

1 1 22 0

0

m mm m

m m m mm m

mm m

mm

m m mm

m

Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-axis.

17: The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (-4, 1). Find the equation of the legs (perpendicular sides) of the triangle.

Solution:

Let ABC be the right angles triangle, where ∠C = 90o There are infinity many such lines. Let m be the slope of AC.

∴ Slope of BC = 1m

Equation of AC: y – 3 = m(x –1) 11 ( 3)x ym

Equation of BC: y – 1 = 1m

(x + 4)

x + 4 = -m(y – 1) For a given value of m, we can get these equations For m = 0, y – 3 = 0; x + 4 = 0 For m → , x – 1 = 0; y – 1 = 0

18: Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Solution:

The equation of the given line is x + 3y = 7 … (1) Let point B (a, b) be the image of point A (3, 8). Accordingly, line (1) is the perpendicular bisector of AB.




Slope of AB = 83

ba

, while the slope of the line (1) = 1

3

Since line (1) is perpendicular to AB, 8 1 13 3

ba

8 13 9

8 3 93 1 ... 2

ba

b aa b

Mid-Point of AB = 3 8,2 2

a b

The mid-point of the line segment AB will also satisfy line (1). Hence, from equation (1), we have

3 83 72 2

a b

3 3 24 143 13 ... 3

a ba b

On solving equations (2) and (3), we obtain a = -1 and b = -4. Thus, the image of the given point with respect to the given line is (-1, -4).

19: If the lines y = 3x + 1 and 2y = x + 3 are equally indicated to the line y = mx + 4, find the value of m.

Solution:

The equation of the given lines are y = 3x + 1 …(1) 2y = x + 3 …(2) y = mx + 4 …(3) Slope of line (1), 1m = 3




Slope of line (2), 212

m

Slope of line (3), 3m = m It is given that lines (1) and (2) are equally inclined to line (3). This means that the given angle between lines (1) and (3) equals the angle between lines (2) and (3).

1 3 2 3

1 3 2 3

2 2

2

2

1 1

13 2

11 3 12

3 1 21 3 23 1 21 3 23 1 2 3 1 21 3 2 1 3 23 1 2If , then1 3 2

3 2 1 2 1 3

6 1 65 5 0

1 0

1, whi

m m m mm m m m

mmm m

m mm m

m mm mm m m morm m m m

m mm m

m m m m

m m m mm

m

m

ch is not realHence, this case is not possible.

2 2

2

3 1 2If ,1 3 23 2 1 2 1 3

6 1 6

7 2 7 0

2 4 4 7 72 7

2 2 1 4914

1 5 27

1 5 2Thus, the required value of m is 7

m m thenm mm m m m

m m m m

m m

m

m

m




20: If sum of the perpendicular distance of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.

Solution:

The equation of the given lines are x + y – 5 = 0 … (1) 3x – 2y + 7 = 0 … (2) The perpendicular distance of P (x, y) from lines (1) and (2) are respectively given by

1 22 2 2 2

1 2

1 2

5 3 2 7

1 1 3 2

5 3 2 7. .,

2 13It is given that 10

5 3 2 710

2 1313 5 2 3 2 7 10 26 0

13 5 2 3 2 7 10 26 0

Assuming 5 and 3 2 7 are positive

13

x y x yd and d

x y x yi e d and d

d dx y x y

x y x y

x y x y

x y x y

x

13 5 13 3 2 2 2 7 2 10 26 0y x y

13 3 2 13 2 2 7 2 5 13 10 26 0x y , which is the equation of a line.

Similarly, we can obtain the equation of line for any signs of (x + y – 5) and (3x – 2y + 7). Thus, point P must move on a line.

21: Find equation of the line which is equidistant from parallel lines 9x + 6y – 7=0 and 3x + 2y + 6 = 0.

Solution:

The equation of the given lines are 9x + 6y – 7=0 …(1) 3x + 2y + 6 = 0 …(2) Let p(h, k) be the arbitrary point is equidistant from lines (1) and (2). The perpendicular distance of P (h, k) from line (1) is given by




1 2 2

2 2 2

1 2

9 6 7 9 6 7 9 6 7117 3 139 6

The perpendicular distance of P (h, k) from line (2) is given by3 2 6 3 2 6

133 2

Since p(h, k) is equidistant from lines (1) and (2), 9 6 7 3 2

3 13

h k h k h kd

h k h kd

d dh k h

613

9 6 7 3 3 2 6

9 6 7 3 3 2 6

9 6 7 3 3 2 6 9 6 7 3 3 2 6

The case 9 6 7 3 3 2 6 is not possible as

9 6 7 3 3 2 6 7 18 which is absurd

9 6 7 3 3 2 69 6 7 9 6 18

18 12

k

h k h k

h k h k

h k h k or h k h k

h k h k

h k h k

h k h kh k h k

h

11 0Thus, the required equation of the line is 18 12 11 0

kx y

22: A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

Solution:

Let the coordinates of point A be (a, 0). Draw a line (AL) perpendicular to the x-axis.




We know that angle of incidence is equal to angle of reflection. Hence, let

Let

180 2 180 2 90

180 180 2

1803-0Now, slope of line AC=5-a

3tan ... 15

2-0SLope of line AB=1-a2tan 180

1-a2tan

1-a2tan ... 2

a-1From equatio

BAL CALCAX

OAB

BAX

a

n (1) and (2), we obtain3 2

5 a-13 3 10 2

135

13Thus, the coordinates of point A are ,0 .5

aa a

a

23: Prove that the product of the lengths of the perpendiculars drawn from the points

2 2 2 2 2x,0 and - ,0 to the line cos sin 1 is b .a

ya b a bb

Solution:

The equation of the given line is x cos sin 1a

yb

Or, bxcos sin 0ay ab … (1)

Length of the perpendicular from point 2 2 ,0a b to line (1) is




2 2 2 2

1 2 2 2 2 2 2 2 2

cos sin 0 cos... 2

cos sin cos sin

b a b a ab b a b abp

b a b a

Length of the perpendicular from point ( 2 2 ,0a b ) to line (2) is

2 2 2 2

2 2 2 2 2 2 2 2 2

cos sin 0 cos...(3

cos sin cos sin

b a b a ab b a b abp

b a b a

On multiplying equations (2) and (3), we obtain

2 2 2 2

1 2 22 2 2 2

2 2 2 2

2 2 2 2

2 22 2

2 2 2 2

2 2 2 2 2 2

2 2 2 2

2 2 2 4 2 2 2

2 2 2 2

2 2 2

cos cos

cos sin

cos cos

cos sin

cos

cos sin

cos

cos sin

cos coscos sin

cos

b a b ab b a b abp p

b a

b a b ab b a b ab

b a

b a b ab

b a

b a b a b

b a

a b b a bb a

b a

2 2 2

2 2 2 2

2 2 2 2 2 2 2 2 22 2

2 2 2 2

2 2 2 2 2

2 2 2 2

2 2 2 2 2

2 2 2 2

2

coscos sin

cos cos sin cossin cos 1

cos sincos sin

cos sincos sin

cos sin

b ab a

b a b a ab a

b b a

b ab b a

b a

b

Hence, proved.

24: A person standing at the junction (crossing) of two straight paths represented by the equation 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 equation of the path that he should follow.

Solution:




The equations of the given lines are 2x– 3y + 4 = 0 … (1) 3x + 4y – 5 = 0 … (2) 6x – 7y + 8 = 0 … (3) The person is standing at the junction of the paths represented by lines (1) and (2).

On solving equations (1) and (2), we obtain 1 22and 17 17

x y

Thus, the person is standing at point 1 22,17 17

The person can reach path (3) in the least time if he walks along the perpendicular line to (3)

from point 1 22,17 17

.

Slope of the line (3) = 67

Slope of the line perpendicular to line (3) = 1 76 67

The equation of the line passing through 1 22,17 17

and having a slope of 76

is given by

22 7 117 6 17

6 17 22 7 17 1102 132 119 7119 102 125

y x

y xy xx y

Hence, the path that the person should follow is 119 102 125x y .




Miscellaneous Exercise Page: 233

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