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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9.

The Class 11 textbook contains an exercise at the end of every chapter. This exercise contains questions that cover all the topics in the chapter. Chapter 10 Straight Lines of Class 11 Maths is categorised under the CBSE Syllabus for 2023-24. The Miscellaneous Exercise of NCERT Solutions for Class 11 Maths Chapter 10- Straight Lines is based on the following topics:

1. Slope of a Line
2. Various Forms of the Equation of a Line
3. General Equation of a Line
4. Distance of a Point From a Line

Mathematics is a subject that needs more and more practice to understand the best problem-solving method. Practising with the help of these NCERT Solutions for Class 11 Maths will help the students in scoring high marks in the examination.

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

### Access Other Exercise Solutions of Class 11 Maths Chapter 10 – Straight Lines

Exercise 10.1 Solutions 14 Questions

Exercise 10.2 Solutions 20 Questions

Exercise 10.3 Solutions 18 Questions

Also explore – NCERT Class 11 Solutions

#### Access Solutions for Class 11 Maths Chapter 10 Miscellaneous Exercise

1. Find the values of k for which the line (k – 3) x – (4 – k2) y + k2 – 7k + 6 = 0 is

(a) Parallel to the x-axis

(b) Parallel to the y-axis

(c) Passing through the origin

Solution:

It is given that

(k – 3) x – (4 – k2y + k2 – 7k + 6 = 0 … (1)

(a) Here, if the line is parallel to the x-axis,

The slope of the line = Slope of the x-axis

It can be written as

(4 – k2y = (k – 3) x + k2 – 7k + 6 = 0

We get, By further calculation,

k – 3 = 0

k = 3

Hence, if the given line is parallel to the x-axis, then the value of k is 3.

(b) Here, if the line is parallel to the y-axis, it is vertical, and the slope will be undefined.

So, the slope of the given line k2 = 4

k = ± 2

Hence, if the given line is parallel to the y-axis, then the value of k is ± 2.

(c) Here, if the line is passing through (0, 0), which is the origin satisfies the given equation of the line.

(k – 3) (0) – (4 – k2) (0) + k2 – 7k + 6 = 0

By further calculation,

k2 – 7k + 6 = 0

Separating the terms,

k2 – 6k – k + 6 = 0

We get

(k – 6) (k – 1) = 0

k = 1 or 6

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

2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line √3x + y + 2 = 0.

Solution: 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and –6, respectively.

Solution:

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

a + b = 1 …… (1)

ab = – 6 …….. (2)

By solving both equations, we get

a = 3 and b = -2 or a = -2 and b = 3

We know that the equation of the line whose intercepts on the a and b axes is Case I – a = 3 and b = – 2

So, the equation of the line is – 2x + 3y + 6 = 0, i.e. 2x – 3y = 6.

Case II – a = -2 and b = 3

So, the equation of the line is 3x – 2y + 6 = 0, i.e. -3x + 2y = 6

Hence, the required equation of the lines are 2x – 3y = 6 and -3x + 2y = 6.

4. What are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units?

Solution:

Consider (0, b) as the point on the y-axis whose distance from line x/3 + y/4 = 1 is 4 units.

It can be written as 4x + 3y – 12 = 0 ……. (1)

By comparing equation (1) to the general equation of line Ax + By + C = 0, we get

A = 4, B = 3 and C = – 12

We know that the perpendicular distance (d) of a line Ax + By + C = 0 from (x1, y1) is written as By cross multiplication

20 = |3b – 12|

We get

20 = ± (3b – 12)

Here, 20 = (3b – 12) or 20 = – (3b – 12)

It can be written as

3b = 20 + 12 or 3b = -20 + 12

So, we get

b = 32/3 or b = -8/3

Hence, the required points are (0, 32/3) and (0, -8/3).

5. Find the perpendicular distance from the origin to the line joining the points Solution:   6. Find the equation of the line parallel to the y-axis drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

Solution:

Here, the equation of any line parallel to the y-axis is of the form,

x = a ……. (1)

Two given lines are

x – 7y + 5 = 0 …… (2)

3x + y = 0 …… (3)

By solving equations (2) and (3), we get

x = -5/22 and y = 15/22

(-5/ 22, 15/22) is the point of intersection of lines (2) and (3)

If the line x = a passes through point (-5/22, 15/22), we get a = -5/22

Hence, the required equation of the line is x = -5/22.

7. Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point where it meets the y-axis.

Solution:

It is given that

x/4 + y/6 = 1

We can write it as

3x + 2y – 12 = 0

So, we get

y = -3/2 x + 6, which is of the form y = mx + c

Here, the slope of the given line = -3/2

So, the slope of line perpendicular to the given line = -1/ (-3/2) = 2/3

Consider the given line intersects the y-axis at (0, y)

By substituting x as zero in the equation of the given line,

y/6 = 1

y = 6

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

We know that the equation of the line that has a slope of 2/3 and passes through the point (0, 6) is

(y – 6) = 2/3 (x – 0)

By further calculation,

3y – 18 = 2x

So, we get

2x – 3y + 18 = 0

Hence, the required equation of the line is 2x – 3y + 18 = 0.

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

Solution:

It is given that

y – x = 0 …… (1)

x + y = 0 …… (2)

x – k = 0 ……. (3)

Here, the point of intersection of

Lines (1) and (2) is

x = 0 and y = 0

Lines (2) and (3) is

x = k and y = – k

Lines (3) and (1) is

x = k and y = k

So, the vertices of the triangle formed by the three given lines are (0, 0), (k, -k) and (k, k)

Here, the area of triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is

½ |x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|

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

= ½ |0 (-k – k) + k (k – 0) + k (0 + k)| square units

By further calculation,

= ½ |k2 + k2| square units

So, we get

= ½ |2k2|

= k2

Hence, the area of the triangle formed by the three given lines is k2.

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:

It is given that

3x + y – 2 = 0 …… (1)

px + 2y – 3 = 0 ….. (2)

2x – y – 3 = 0 …… (3)

By solving equations (1) and (3), we get

x = 1 and y = -1

Here, the three lines intersect at one point, and the point of intersection of lines (1) and (3) will also satisfy line (2)

p (1) + 2 (-1) – 3 = 0

By further calculation,

p – 2 – 3 = 0

So, we get

p = 5

Hence, the required value of p is 5.

10. If three lines whose equations are y = m1x + c1, y = m2x + c2 and y = m3x + c3 are concurrent, then show that m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0.

Solution:

It is given that

y = m1x + c1 ….. (1)

y = m2x + c2 ….. (2)

y = m3x + c3 ….. (3)

By subtracting equation (1) from (2), we get

0 = (m2 – m1) x + (c2 – c1)

(m1 – m2) x = c2 – c1

So, we get  Taking out the common terms

m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0

Therefore, m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0.

11. Find the equation of the lines through the point (3, 2), which makes an angle of 45° with the line x –2y = 3.

Solution:

Consider m1 as the slope of the required line

It can be written as

y = 1/2 x – 3/2 which is of the form y = mx + c

So, the slope of the given line m2 = 1/2

We know that the angle between the required line and line x – 2y = 3 is 45o

If θ is the acute angle between lines l1 and l2 with slopes m1 and m2,  It can be written as

2 + m1 = 1 – 2m1 or 2 + m1 = -1 + 2m1

m1 = -1/3 or m1 = 3

Case I – m1 = 3

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

y – 2 = 3 (x – 3)

By further calculation,

y – 2 = 3x – 9

So, we get

3x – y = 7

Case II – m1 = -1/3

Here, the equation of the line passing through (3, 2) and having a slope of -1/3 is

y – 2 = -1/3 (x – 3)

By further calculation,

3y – 6 = – x + 3

So, we get

x + 3y = 9

Hence, the equations of the lines are 3x – y = 7, and x + 3y = 9.

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

Solution:

Consider the equation of the line having equal intercepts on the axes as

x/a + y/a = 1

It can be written as

x + y = a ….. (1)

By solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we get

x = 1/13 and y = 5/13

(1/13, 5/13) is the point of intersection of two given lines.

We know that equation (1) passes through the point (1/13, 5/13)

1/13 + 5/13 = a

a = 6/13

So, equation (1) passes through (1/13, 5/13).

1/13 + 5/13 = a

We get

a = 6/13

Here, equation (1) becomes

x + y = 6/13

13x + 13y = 6

Hence, the required equation of the line is 13x + 13y = 6.

13. Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c is .

Solution:

Consider y = m1x as the equation of the line passing through the origin   14. In what ratio the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?

Solution:  By cross multiplication

– k + 5 = 1 + k

We get

2k = 4

k = 2

Hence, the line joining the points (-1, 1) and (5, 7) is divided by the 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:

It is given that

2x – y = 0 ….. (1)

4x + 7y + 5 = 0 …… (2)

Here A (1, 2) is a point on line (1)

Consider B as the point of intersection of lines (1) and (2) By solving equations (1) and (2), we get x = -5/18 and y = – 5/9

So, the coordinates of point B is (-5/18, -5/9)

From the distance formula, the distance between A and B  Hence, the required distance is .

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.

Solution:

Consider y = mx + c as the line passing through the point (-1, 2)

So, we get

2 = m (-1) + c

By further calculation,

2 = -m + c

c = m + 2

Substituting the value of c,

y = mx + m + 2 …… (1)

So, the given line is

x + y = 4 ……. (2)

By solving both equations, we get By cross multiplication

1 + m2 = m2 + 1 + 2m

So, we get

2m = 0

m = 0

Hence, 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 points (1, 3) and (−4, 1). Find the equation of the legs (perpendicular sides) of the triangle.

Solution:

Consider ABC as the right angles triangle where ∠C = 90o

Here, infinitely such lines are present.

m is the slope of AC

So, the slope of BC = -1/m

Equation of AC,

y – 3 = m (x – 1)

By cross multiplication,

x – 1 = 1/m (y – 3)

Equation of BC,

y – 1 = – 1/m (x + 4)

By cross multiplication,

x + 4 = -m (y – 1)

By considering the values of m, we get

If m = 0,

So, we get

y – 3 = 0, x + 4 = 0

If m = ∞,

So, we get

x – 1 = 0, y – 1 = 0 we get x = 1, y = 1

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:

It is given that

x + 3y = 7 ….. (1)

Consider B (a, b) as the image of point A (3, 8)

So, line (1) is the perpendicular bisector of AB.   On further simplification,

a + 3b = – 13 ….. (3)

By solving equations (2) and (3), we get

a = -1 and b = -4

Hence, 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 inclined to the line y = mx + 4, find the value of m.

Solution:

It is given that

y = 3x + 1 …… (1)

2y = x + 3 …… (2)

y = mx + 4 …… (3)

Here, the slopes of

Line (1), m1 = 3

Line (2), m2 = ½

Line (3), m3 = m

We know that lines (1) and (2) are equally inclined to the line (3), which means that the angle between lines (1) and (3) equals the angle between lines (2) and (3). On further calculation,

– m2 + m + 6 = 1 + m – 6m2

So we get

5m2 + 5 = 0

Dividing the equation by 5,

m2 + 1 = 0

m = √-1, which is not real.

Therefore, this case is not possible.

If 20. If the sum of the perpendicular distances 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:  In the same way, we can find the equation of the line for any signs of (x + y – 5) and (3x – 2y + 7).

Hence, point P must move on a line.

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

Solution:  Here,

9h + 6k – 7 = 3 (3h + 2k + 6) or 9h + 6k – 7 = – 3 (3h + 2k + 6)

9h + 6k – 7 = 3 (3h + 2k + 6) is not possible as

9h + 6k – 7 = 3 (3h + 2k + 6)

By further calculation,

– 7 = 18 (is not correct)

We know that

9h + 6k – 7 = -3 (3h + 2k + 6)

By multiplication,

9h + 6k – 7 = -9h – 6k – 18

We get

18h + 12k + 11 = 0

Hence, the required equation of the line is 18x + 12y + 11 = 0.

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: Consider the coordinates of point A as (a, 0)

Construct a line (AL) which is perpendicular to the x-axis

Here, the angle of incidence is equal to the angle of reflection

∠BAL = ∠CAL = Φ

∠CAX = θ

It can be written as

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

On further calculation,

= 180° – θ – 180° + 2θ

θ

So, we get

∠BAX = 180° – θ By cross multiplication,

3a – 3 = 10 – 2a

We get

a = 13/5

Hence, the coordinates of point A are (13/5, 0).

23. Prove that the product of the lengths of the perpendiculars drawn from the points to the line .

Solution:

It is given that We can write it as

bx cos θ + ay sin θ – ab = 0 ….. (1)   24. A person standing at the junction (crossing) 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.

Solution:

It is given that

2x – 3y + 4 = 0 …… (1)

3x + 4y – 5 = 0 ……. (2)

6x – 7y + 8 = 0 …… (3)

Here, the person is standing at the junction of the paths represented by lines (1) and (2).

By solving equations (1) and (2), we get

x = – 1/17 and y = 22/17

Hence, the person is standing at point (-1/17, 22/17).

We know that the person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point (-1/17, 22/17).

Here, the slope of the line (3) = 6/7

We get the slope of the line perpendicular to the line (3) = -1/ (6/7) = – 7/6

So, the equation of the line passing through (-1/17, 22/17) and having a slope of -7/6 is written as By further calculation,

6 (17y – 22) = – 7 (17x + 1)

By multiplication,

102y – 132 = – 119x – 7

We get

1119x + 102y = 125

Therefore, the path that the person should follow is 119x + 102y = 125.