## NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines Miscellaneous Exercise

#### Page No 233:

#### Question 1:

Find the values of

*k*for which the lineis
(a) Parallel to the

*x*-axis,
(b) Parallel to the

*y*-axis,
(c) Passing through the origin.

#### Answer:

The given equation of line is

(

*k*– 3)*x*– (4 –*k*^{2})*y*+*k*^{2}– 7*k*+ 6 = 0 … (1)
(a) If the given line is parallel to the

*x*-axis, then
Slope of the given line = Slope of the

*x*-axis
The given line can be written as

(4 –

*k*^{2})*y*= (*k*– 3)*x*+*k*^{2}– 7*k*+ 6 = 0
, which is of the form

*y*=*mx*+*c*.
∴Slope of the given line =

Slope of the

*x*-axis = 0
Thus, if the given line is parallel to the

*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.

Now, is undefined at

*k*^{2}= 4*k*

^{2}= 4

⇒

*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 equation of line.

Thus, if the given line is passing through the origin, then the value of

*k*is either 1 or 6.#### Question 2:

Find the values of

*θ*and*p*, if the equation**is the normal form of the line.**#### Answer:

The equation of the given line is.

This equation can be reduced as

On dividing both sides by, we obtain

On comparing equation (1) to, we obtain

Since the values of sin

*θ*and cos*θ*are negative,
Thus, the respective values of

*θ*and*p*are and 1#### Question 3:

Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and –6, respectively.

#### Answer:

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**Case I:**

*a*= 3 and

*b*= –2

In this case, the equation of the line is –2

*x*+ 3*y*+ 6 = 0, i.e., 2*x*– 3*y*= 6.**Case II:**

*a*= –2 and

*b*= 3

In this case, the equation of the line is 3

*x*– 2*y*+ 6 = 0, i.e., –3*x*+ 2*y*= 6.
Thus, the required equation of the lines are 2

*x*– 3*y*= 6 and –3*x*+ 2*y*= 6.#### Question 4:

What are the points on the

*y*-axis whose distance from the line is 4 units.#### Answer:

Let (0,

*b*) be the point on the*y*-axis whose distance from line is 4 units.
The given line can be written as 4

*x*+ 3*y*– 12 = 0 … (1)
On comparing equation (1) to the general equation of line

*Ax*+*By*+*C*= 0, we obtain*A*= 4,*B*= 3, and*C*= –12.
It is known that the perpendicular distance (

*d*) of a line*Ax*+*By*+*C*= 0 from a point (*x*_{1},*y*_{1}) is given by.
Therefore, if (0,

*b*) is the point on the*y*-axis whose distance from line is 4 units, then:
Thus, the required points are and.

#### Question 5:

Find the perpendicular distance from the origin to the line joining the points

#### Answer:

The equation of the line joining the points is given by

It is known that the perpendicular distance (

*d*) of a line*Ax*+*By*+*C*= 0 from a point (*x*_{1},*y*_{1}) is given by.
Therefore, the perpendicular distance (

*d*) of the given line from point (*x*_{1},*y*_{1}) = (0, 0) is#### Question 6:

Find the equation of the line parallel to

*y*-axis and drawn through the point of intersection of the lines*x*– 7*y*+ 5 = 0 and 3*x*+*y*= 0.#### Answer:

The equation of any line parallel to the

*y*-axis is of the form*x*=

*a*… (1)

The two given lines are

*x*– 7

*y*+ 5 = 0 … (2)

3

*x*+*y*= 0 … (3)
On solving equations (2) and (3), we obtain.

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

Since line

*x*=*a*passes through point, .
Thus, the required equation of the line is.

#### Question 7:

Find the equation of a line drawn perpendicular to the line through the point, where it meets the

*y*-axis.#### Answer:

The equation of the given line is.

This equation can also be written as 3

*x*+ 2*y*– 12 = 0
, which is of the form

*y*=*mx*+*c*
∴Slope of the given line

∴Slope of line perpendicular to the given line

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
∴The given line intersects the

*y*-axis at (0, 6).
The equation of the line that has a slope of and passes through point (0, 6) is

Thus, the required equation of the line is.

#### Question 8:

Find the area of the triangle formed by the lines

*y*–*x*= 0,*x*+*y*= 0 and*x*–*k*= 0.#### Answer:

The equations 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 (

*x*_{1},*y*_{1}), (*x*_{2},*y*_{2}), and (*x*_{3},*y*_{3}) is.
Therefore, area of the triangle formed by the three given lines

#### Question 9:

Find the value of

*p*so that the three lines 3*x*+*y*– 2 = 0,*px*+ 2*y*– 3 = 0 and 2*x*–*y*– 3 = 0 may intersect at one point.#### Answer:

The equations of the given lines are

3

*x*+*y*– 2 = 0 … (1)*px*+ 2

*y*– 3 = 0 … (2)

2

*x*–*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.#### Question 10:

If three lines whose equations are

concurrent, then show that

#### Answer:

The equations of the given lines are

*y*=

*m*

_{1}

*x*+

*c*

_{1}… (1)

*y*=

*m*

_{2}

*x*+

*c*

_{2}… (2)

*y*=

*m*

_{3}

*x*+

*c*

_{3}… (3)

On subtracting equation (1) from (2), we obtain

On substituting this value of

*x*in (1), we obtain
is the point of intersection of lines (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).

Hence,

#### Question 11:

Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line

*x*–2*y*= 3.#### Answer:

Let the slope of the required line be

*m*_{1}.
The given line can be written as , which is of the form

*y*=*mx*+*c*
∴Slope of the given line =

It is given that the angle between the required line and line

*x*– 2*y*= 3 is 45°.
We know that if

*θ*isthe acute angle between lines*l*_{1}and*l*_{2}with slopes*m*_{1}and*m*_{2 }respectively, then.**Case I:**

*m*

_{1}= 3

The equation of the line passing through (3, 2) and having a slope of 3 is:

*y*– 2 = 3 (

*x*– 3)

*y*– 2 = 3

*x*– 9

3

*x*–*y*= 7**Case II:**

*m*

_{1}=

The equation of the line passing through (3, 2) and having a slope of is:

Thus, the equations of the lines are 3

*x*–*y*= 7 and*x*+ 3*y*= 9.#### Question 12:

Find the equation of the line passing through the point of intersection of the lines 4

*x*+ 7*y*– 3 = 0 and 2*x*– 3*y*+ 1 = 0 that has equal intercepts on the axes.#### Answer:

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

On solving equations 4

*x*+ 7*y*– 3 = 0 and 2*x*– 3*y*+ 1 = 0, we obtain.
is the point of intersection of the two given lines.

Since equation (1) passes through point,

∴ Equation (1) becomes

Thus, the required equation of the line is.

#### Page No 234:

#### Question 13:

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

*θ*with the line.#### Answer:

Let the equation of the line passing through the origin be

*y*=*m*_{1}*x*.
If this line makes an angle of

*θ*with line*y*=*mx*+*c*, then angle*θ*is given by**Case I:**

**Case II:**

Therefore, the required line is given by.

#### Question 14:

In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line

*x*+

*y*= 4?

#### Answer:

The equation of the line joining the points (–1, 1) and (5, 7) is given by

The equation of the given line is

*x*+

*y*– 4 = 0 … (2)

The point of intersection of lines (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,
Thus, the line joining the points (–1, 1) and (5, 7) is divided by line

*x*+

*y*= 4 in the ratio 1:2.

#### Question 15:

Find the distance of the line 4

*x*+ 7*y*+ 5 = 0 from the point (1, 2) along the line 2*x*–*y*= 0.#### Answer:

The given lines are

2

*x*–*y*= 0 … (1)
4

*x*+ 7*y*+ 5 = 0 … (2)
A (1, 2) is a point on line (1).

Let B be the point of intersection of lines (1) and (2).

On solving equations (1) and (2), we obtain.

∴Coordinates of point B are.

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

Thus, the required distance is.

#### Question 16:

Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line

*x*+*y*= 4 may be at a distance of 3 units from this point.#### Answer:

Let

*y*=*mx*+*c*be the line through point (–1, 2).
Accordingly, 2 =

*m*(–1) +*c*.
⇒ 2 = –

*m*+*c*
⇒

*c*=*m*+ 2
∴

*y*=*mx*+*m*+ 2 … (1)
The given line is

*x*+

*y*= 4 … (2)

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

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

Since this point is at a distance of 3 units from point (– 1, 2), according to distance formula,

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

*x*-axis.#### Question 18:

Find the image of the point (3, 8) with respect to the line

*x*+ 3*y*= 7 assuming the line to be a plane mirror.#### Answer:

The equation of the given line is

*x*+ 3

*y*= 7 … (1)

Let point B (

*a*,*b*) be the image of point A (3, 8).
Accordingly, line (1) is the perpendicular bisector of AB.

Since line (1) is perpendicular to AB,

The mid-point of line segment AB will also satisfy line (1).

Hence, from equation (1), we have

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).

#### Question 19:

If the lines

*y*= 3*x*+ 1 and 2*y*=*x*+ 3 are equally inclined to the line*y*=*mx*+ 4, find the value of*m*.#### Answer:

The equations of the given lines are

*y*= 3

*x*+ 1 … (1)

2

*y*=*x*+ 3 … (2)*y*=

*mx*+ 4 … (3)

Slope of line (1),

*m*_{1}= 3
Slope of line (2),

Slope of line (3),

*m*_{3 }=*m*
It is given that lines (1) and (2) are equally inclined to line (3). This means that

the angle between lines (1) and (3) equals the angle between lines (2) and (3).

Thus, the required value of

*m*is.#### Question 20:

If sum of the perpendicular distances of a variable point P (

*x*,*y*) from the lines*x*+*y*– 5 = 0 and 3*x*– 2*y*+ 7 = 0 is always 10. Show that P must move on a line.#### Answer:

The equations of the given lines are

*x*+

*y*– 5 = 0 … (1)

3

*x*– 2*y*+ 7 = 0 … (2)
The perpendicular distances of P (

*x*,*y*) from lines (1) and (2) are respectively given by
It is given that.

, which is the equation of a line.

Similarly, we can obtain the equation of line for any signs of.

Thus, point P must move on a line.

#### Question 21:

Find equation of the line which is equidistant from parallel lines 9

*x*+ 6*y*– 7 = 0 and 3*x*+ 2*y*+ 6 = 0.#### Answer:

The equations of the given lines are

9

*x*+ 6*y*– 7 = 0 … (1)
3

*x*+ 2*y*+ 6 = 0 … (2)
Let P (

*h*,*k*) be the arbitrary point that is equidistant from lines (1) and (2). The perpendicular distance of P (*h*,*k*) from line (1) is given by
The perpendicular distance of P (

*h,**k*) from line (2) is given by
Since P (

*h*,*k*) is equidistant from lines (1) and (2),
∴

9

*h*+ 6*k*– 7 = – 9*h*– 6*k*– 18
⇒ 18

*h*+ 12*k*+ 11 = 0
Thus, the required equation of the line is 18

*x*+ 12*y*+ 11 = 0.#### Question 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.#### Answer:

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

∠BAL = ∠CAL =

*Φ*
Let ∠CAX =

*θ*
∴∠OAB = 180° – (

*θ*+ 2*Φ*) = 180° – [*θ*+ 2(90° –*θ*)]
= 180° –

*θ*– 180° + 2*θ*
=

*θ*
∴∠BAX = 180° –

*θ*
From equations (1) and (2), we obtain

Thus, the coordinates of point A are.

#### Question 23:

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

#### Answer:

The equation of the given line is

Length of the perpendicular from point to line (1) is

Length of the perpendicular from point to line (2) is

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

Hence, proved.

#### Question 24:

A person standing at the junction (crossing) of two straight paths represented by the equations 2

*x*– 3*y*+ 4 = 0 and 3*x*+ 4*y*– 5 = 0 wants to reach the path whose equation is 6*x*– 7*y*+ 8 = 0 in the least time. Find equation of the path that he should follow.#### Answer:

The equations of the given lines are

2

*x*– 3*y*+ 4 = 0 … (1)
3

*x*+ 4*y*– 5 = 0 … (2)
6

*x*– 7*y*+ 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.

Thus, the person is standing at point.

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

∴Slope of the line perpendicular to line (3)

The equation of the line passing through and having a slope of is given by

Hence, the path that the person should follow is.

#### Question 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.

#### Answer:

Let A(1,3) and B(−4,1) be the coordinates of the end points of the hypotenuse.

Now, plotting the line segment joining the points A(1,3) and B(−4,1) on the coordinate plane, we will get two right triangles with AB as the hypotenuse. Now from the diagram, it is clear that the point of intersection of the other two legs of the right triangle having AB as the hypotenuse can be either P or Q.

CASE 1: When âˆ† APB is taken. The perpendicular sides in âˆ† APB are AP and PB. Now, side PB is parallel to x-axis and at a distance of 1 units above x-axis. So, equation of PB is,

*y*=1 or*y*−1=0. The side AP is parallel to y-axis and at a distance of 1 units on the right of y-axis. So, equation of AP is*x*=1 or*x*−1=0.
CASE 2: When âˆ† AQB is taken. The perpendicular sides in âˆ† AQB are AQ and QB. Now, side AQ is parallel to x-axis and at a distance of 3 units above x-axis. So, equation of AQ is,

*y*=3 or*y*−3=0. The side QB is parallel to y-axis and at a distance of 4 units on the left of y-axis. So, equation of QB is*x*=−4 or*x*+4=0.
Hence, the equation of the legs are :

*x*=1,*y*=1 or*x*=−4,*y*=3_{}

^{}