 # NCERT Solutions for Class 11 Maths Chapter 10 - Straight Lines Exercise 10.1

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

Chapter 10 Straight Lines of Class 11 Maths is categorised under the latest CBSE Syllabus for the session 2023-24. The answers to the questions given in the first exercise of Chapter 10, NCERT Class 11 Textbooks are provided here. Exercise 10.1 of NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines is based on the following topics:

1. Introduction to Straight Lines
2. The slope of a Line
1. The slope of a line when the coordinates of any two points on the line are given
2. Conditions for parallelism and perpendicularity of lines in terms of their slopes
3. The angle between two lines
4. Collinearity of three points

For students to score high marks in the board exam, practising with these NCERT Solutions is mandatory. These NCERT Solutions of Class 11 Maths are helpful for students to score well in the CBSE board examination, as they work for them as reference tools to do the revision.

## NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines Exercise 10.1

### Access Other Exercise Solutions of Class 11 Maths Chapter 10

Exercise 10.2 Solutions 20 Questions

Exercise 10.3 Solutions 18 Questions

Miscellaneous Exercise on Chapter 10 Solutions 24 Questions

Also explore – NCERT Class 11 Solutions

#### Access Solutions for Class 11 Maths Chapter 10.1 Exercise

1. Draw a quadrilateral in the Cartesian plane whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

Solution:

Let ABCD be the given quadrilateral with vertices A (-4,5) , B (0,7), C (5.-5) and D (-4,-2).

Now, let us plot the points on the Cartesian plane by joining the points AB, BC, CD, AD, which gives us the required quadrilateral. To find the area, draw the diagonal AC.

So, area (ABCD) = area (∆ABC) + area (∆ADC)

Then, area of triangle with vertices (x1,y1) , (x2, y2) and (x3,y3) is

Are of ∆ ABC = ½ [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]

= ½ [-4 (7 + 5) + 0 (-5 – 5) + 5 (5 – 7)] unit2

= ½ [-4 (12) + 5 (-2)] unit2

= ½ (58) unit2

= 29 unit2

Are of ∆ ACD = ½ [x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]

= ½ [-4 (-5 + 2) + 5 (-2 – 5) + (-4) (5 – (-5))] unit2

= ½ [-4 (-3) + 5 (-7) – 4 (10)] unit2

= ½ (-63) unit2

= -63/2 unit2

Since area cannot be negative, area ∆ ACD = 63/2 unit2

Area (ABCD) = 29 + 63/2

= 121/2 unit2

2. The base of an equilateral triangle with side 2a lies along the y-axis, such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Solution: Let us consider ABC to be the given equilateral triangle with side 2a.

Where, AB = BC = AC = 2a

In the above figure, assuming that the base BC lies on the x-axis, such that the mid-point of BC is at the origin, i.e., BO = OC = a, where O is the origin.

The coordinates of point C are (0, a), and that of B is (0,-a).

The line joining a vertex of an equilateral ∆ with the mid-point of its opposite side is perpendicular.

So, vertex A lies on the y –axis.

By applying Pythagoras’ theorem,

(AC)2 = OA2 + OC2

(2a)2= a2 + OC2

4a2 – a2 = OC2

3a2 = OC2

OC =√3a

Co-ordinates of point C = ± √3a, 0

∴ The vertices of the given equilateral triangle are (0, a), (0, -a), (√3a, 0)

Or (0, a), (0, -a) and (-√3a, 0)

3. Find the distance between P (x1, y1) and Q (x2, y2) when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Solution:

Given:

Points P (x1, y1) and Q(x2, y2)

(i) When PQ is parallel to y-axis, then x1 = x2

So, the distance between P and Q is given by,  = |y2 – y1|

(ii) When PQ is parallel to the x-axis, then y1 = y2

So, the distance between P and Q is given by,

= = = |x2 – x1|

4. Find a point on the x-axis which is equidistant from points (7, 6) and (3, 4).

Solution:

Let us consider (a, 0) to be the point on the x-axis that is equidistant from the point (7, 6) and (3, 4).

So, Now, let us square on both sides we get,

a2 – 14a + 85 = a2 – 6a + 25

-8a = -60

a = 60/8

= 15/2

∴ The required point is (15/2, 0).

5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

Solution:

The co-ordinates of the mid-point of the line segment joining the points P (0, – 4) and B (8, 0) are (0+8)/2, (-4+0)/2 = (4, -2)

The slope ‘m’ of the line non-vertical line passing through the point (x1, y1) and

(x2, y2) is given by m = (y2 – y1)/(x2 – x1) where, x ≠ x1

The slope of the line passing through (0, 0) and (4, -2) is (-2-0)/(4-0) = -1/2

∴ The required slope is -1/2.

6. Without using Pythagoras’ theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right-angled triangle.

Solution:

The vertices of the given triangle are (4, 4), (3, 5) and (–1, –1).

The slope (m) of the line non-vertical line passing through the point (x1, y1) and

(x2, y2) is given by m = (y2 – y1)/(x2 – x1) where, x ≠ x1

So, the slope of the line AB (m1) = (5-4)/(3-4) = 1/-1 = -1

The slope of the line BC (m2) = (-1-5)/(-1-3) = -6/-4 = 3/2

The slope of the line CA (m3) = (4+1)/(4+1) = 5/5 = 1

It is observed that m1.m3 = -1.1 = -1

Hence, the lines AB and CA are perpendicular to each other.

∴ Given triangle is right-angled at A (4, 4).

And the vertices of the right-angled ∆ are (4, 4), (3, 5) and (-1, -1).

7. Find the slope of the line, which makes an angle of 30° with the positive direction of the y-axis measured anticlockwise.

Solution:

We know that if a line makes an angle of 30° with the positive direction of the y-axis measured anti-clock-wise, then the angle made by the line with the positive direction of the x-axis measures anti-clock-wise is 90° + 30° = 120°

∴ The slope of the given line is tan 120° = tan (180° – 60°)

= – tan 60°

= –3

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

Solution:

If the points (x, – 1), (2, 1) and (4, 5) are collinear, then the Slope of AB = Slope of BC

Then, (1+1)/(2-x) = (5-1)/(4-2)

2/(2-x) = 4/2

2/(2-x) = 2

2 = 2(2-x)

2 = 4 – 2x

2x = 4 – 2

2x = 2

x = 2/2

= 1

∴ The required value of x is 1.

9. Without using the distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

Solution: Let the given point be A (-2, -1) , B (4, 0) , C ( 3, 3) and D ( -3, 2).

So now, the slope of AB = (0+1)/(4+2) = 1/6

The slope of CD = (3-2)/(3+3) = 1/6

Hence, the Slope of AB = the Slope of CD

∴ AB ∥ CD

Now,

The slope of BC = (3-0)/(3-4) = 3/-1 = -3

The slope of AD = (2+1)/(-3+2) = 3/-1 = -3

Hence, the Slope of BC = Slope of AD

Thus, the pair of opposite sides are quadrilateral are parallel; so, we can say that ABCD is a parallelogram.

Hence, the given vertices, A (-2, -1), B (4, 0), C(3, 3) and D(-3, 2) are vertices of a parallelogram.

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

Solution:

The Slope of the line joining the points (3, -1) and (4, -2) is given by,

m = (y2 – y1)/(x2 – x1) where, x ≠ x1

m = (-2 –(-1))/(4-3)

= (-2+1)/(4-3)

= -1/1

= -1

The angle of inclination of the line joining the points (3, -1) and (4, -2) is given by,

tan θ = -1

θ = (90° + 45°) = 135°

∴ The angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.

11. The slope of a line is double the slope of another line. If the tangent of the angle between them is 1/3, find the slopes of the lines.

Solution:

Let us consider ‘m1’ and ‘m’ to be the slope of the two given lines, such that m1 = 2m

We know that if θ is the angle between the lines l1 and l2 with slope m1 and m2, then 1+2m2 = -3m

2m2 +1 +3m = 0

2m (m+1) + 1(m+1) = 0

(2m+1) (m+1)= 0

m = -1 or -1/2

If m = -1, then the slope of the lines are -1 and -2.

If m = -1/2, then the slope of the lines are -1/2 and -1.

Case 2: 2m2 – 3m + 1 = 0

2m2 – 2m – m + 1 = 0

2m (m – 1) – 1(m – 1) = 0

m = 1 or 1/2

If m = 1, then the slope of the lines are 1 and 2.

If m = 1/2, then the slope of the lines are 1/2 and 1.

∴ The slope of the lines are [-1 and -2] or [-1/2 and -1] or [1 and 2] or [1/2 and 1]

12. A line passes through (x1, y1) and (h, k). If the slope of the line is m, show that k – y1 = m (h – x1).

Solution:

Given, the slope of the line is ‘m’.

The slope of the line passing through (x1, y1) and (h, k) is (k – y1)/(h – x1).

So,

(k – y1)/(h – x1)  = m

(k – y1) = m (h – x1)

Hence, proved.

13. If three points (h, 0), (a, b) and (0, k) lie on a line, show that a/h + b/k = 1

Solution:

Let us consider if the given points A (h, 0), B (a, b) and C (0, k) lie on a line.

Then, the slope of AB = slope of BC

(b – 0)/(a – h) = (k – b)/(0 – a)

Let us simplify, and we get

-ab = (k-b) (a-h)

-ab = ka- kh –ab +bh

ka +bh = kh

Divide both sides by kh, and we get

ka/kh + bh/kh = kh/kh

a/h + b/k = 1

Hence, proved.

14. Consider the following population and year graph (Fig 10.10), find the slope of the line AB and using it, find what will be the population in the year 2010? Solution:

We know that line AB passes through points A (1985, 92) and B (1995, 97).

Its slope will be (97 – 92)/(1995 – 1985) = 5/10 = 1/2

Let ‘y’ be the population in the year 2010. Then, according to the given graph, AB must pass through point C (2010, y).

So now, slope of AB = slope of BC 15/2 = y – 97

y = 7.5 + 97 = 104.5

∴ The slope of line AB is 1/2, while in the year 2010, the population will be 104.5 crores.