NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines.

Board	CBSE
Textbook	NCERT
Class	Class 11
Subject	Maths
Chapter	Chapter 10
Chapter Name	Straight Lines
Exercise	Ex 10.1, Ex 10.2, Ex 10.3
Number of Questions Solved	52
Category	NCERT Solutions

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

Chapter 10 Straight Lines Exercise 10.1

Question 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:
The figure of quadrilateral whose vertices are A(- 4, 5), B(0, 7), C(5, -5) and D(-4, -2) is shown in the below figure.

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

Question 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:
We are given that co-ordinates of P is (x₁, y₁) and Q is (x₂, y₁).
Distance between the points P(x₁, y₁) and Q(x₂, y₁) is

Question 4.
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Solution:
Let the point be P(x, y). Since it lies on the x-axis ∴ y = 0 i.e., required point be (x, 0).
Since the required point is equidistant from points A(7, 6) and B(3, 4) ⇒ PA = PB

Question 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:
We are given that P(0, -4) and B(8, 0).
Let A be the midpoint of PB, then

Question 6.
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle.
Solution:
Let A(4, 4), B(3, 5) and C(-1, -1) be the vertices of ∆ABC.
Let m₁ and m₂ be the slopes of AB and AC respectively.

Question 7.
Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Solution:
The given line makes an angle of 90° + 30° = 120° with the positive direction of x-axis.
Hence, m = tan 120° = – .

Question 8.
Find the value of x for which the points (x, -1), (2, 1) and (4,5) are collinear.
Solution:
Let A(x, -1), B(2, 1) and C(4, 5) be the given collinear points. Then by collinearity of A, B, C, we have slope of AB = slope of BC

Question 9.
Without using distance formula, show that points (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.
Solution:
Let A(-2, -1), B(4, 0), C(3, 3) and D(-3, 2) be the vertices of the given quadrilateral ABCD. Then,

Question 10.
Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2).
Solution:
We are given that the points are A(3, -1) and B(4, -2)

Question 11.
The slope of a line is double of the slope of another line. If tangent of the angle between them is , find the slopes of the lines.
Solution:
Let m₁ and m₂ be the slopes of two lines.

Question 12.
A line passes through (x₁, y₂) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).
Solution:
A line passes through (x₁, y₁) and (h, k). Also, the slope of the line is m.

Question 13.
If three points (h, 0), (a, b) and (0, k) lie on a line, show that
Solution:
Let A(h, 0), B(o, b) and C(0, k) be the given collinear points.
∴ Slope of AB = Slope of BC

Question 14.
Consider the following population and year graph, find the slope of the line AB and using it, find what will be the population in the year 2010?
Solution:
Slope of AB +
Let the population in year 2010 is y, and co-ordinate of C is (2010, y) then, slope of AB = slope of BC

Chapter 10 Straight Lines Exercise 10.2

In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:

Question 1.
Write the equations for the x-and y-axes.
Solution:
We know that the ordinate of each point on the x-axis is 0.
If P(x, y) is any point on the x-axis, then y = 0.
∴ Equation of x-axis is y = 0.
Also, we know that the abscissa of each point on the y-axis is 0. If P(x, y) is any point on the y-axis, then x = 0.
∴ Equation of y-axis is x = 0.

Question 2.
Passing through the point (-4,3) with slope .
Solution:
We know that the equation of a line with slope m and passing through the point (x₀, y₀) is given by (y – y₀) = m (x – x₀).

Question 3.
Passing through (0, 0) with slope m.
Solution:
We know that the equation of a line with slope m and passing through the point (x₀, y₀) is given by (y – y₀) = m(x – x₀)
Here, slope = m, x₀ = 0, y₀ = 0 Required equation is (y – 0) = m(x – 0)
⇒ y = mx.

Question 4.
Passing through (2,2^3) and inclined with the x-axis at an angle of 75°.
Solution:
We know that the equation of a line with slope m and passing through the point (X₀, y₀) is given by (y – y₀) = m(x – x₀)

Question 5.
Intersecting the x-axis at a distance of 3 units to the left of origin with slope -2.
Solution:
We know that the equation of a line with slope m and passing through the point
(x₀, y₀) is given by (y – y₀) = m(x – x₀).
Here, m = – 2, x₀ = – 3, y₀ = 0
y-0 = -2(x + 3) ⇒ 2x + y + 6 = 0

Question 6.
Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.
Solution:
We know that the equation of line with slope m and passing through the point (x₀, y₀) is given by (y – y₀) = m(x – x₀)

Question 7.
Passing through the points (-1,1) and (2, -4).
Solution:
Let the given points be A(-1, 1) and B(2, -4).
We know that the equation of a line passing through the given points (xx, y,) and (x2, y2) is given by

Question 8.
Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.
Solution:
Here, we are given that p = 5 and ⍵ = 30°.

Question 9.
The vertices of ∆PQR are P(2, 1), Q(-2, 3) and ff(4, 5). Find equation of the median through the vertex R.
Solution:
The vertices of ∆PQR are P( 2, 1), Q(-2, 3) and R(4, 5).
Let S be the midpoint of PQ.

Question 10.
Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2,5) and (-3,6).
Solution:
Let M(2, 5) and N(-3, 6) be the end points of the given line segment.

Question 11.
A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.
Solution:
Let A(1, 0) and B( 2, 3) be the given points and D divides the line segment in the ratio 1 : n.

Question 12.
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2,3).
Solution:
Let the required line make intercepts a on the x-axis and y-axis.
Then its equation is
⇒ x + y = a … (i)
Since (i) passes through the point (2, 3), we have
2 + 3 = a ⇒ a = 5
So, required equation of the line is
⇒ x + y = 5.

Question 13.
Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Solution:
Let the intercepts made by the line on the x-axis and y-axis be o and 9 – a respectively.
Then its equation is

Since it passes through point (2, 2), we have
⇒ 2(9 – a) + 2a = a(9 – a)
⇒ 18 – 2a + 2a = 9a – 9a²
⇒ 18 = 9a – a² v a² – 9a + 18 = 0
⇒ a² – 6a – 3a + 18 = 0
⇒ a(a – 6) – 3 (a – 6) = 0 ⇒ a = 3, 6
Now, if a = 3 ⇒ b = 9 – 3 = 6 and if a = 6 ⇒ b = 9 – 6 = 3
So, required equation is

i.e., 2x + y – 6 = 0 or x + 2y – 6 = 0.

Question 14.
Find equation of the line through the point (0, 2) making an angle with the positive x-axis. Also, find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Solution:
Here, m = tan =
The equation of the line passing through point (0, 2) is y -2 = (x – 0)
⇒ + y – 2 = 0
The slope of line parallel to
+ y – 2 = 0 is .
Since, it passes through (0, -2).
So, the equation of line is
y + 2= (x – 0)
⇒ + y + 2 = 0.

Question 15.
The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.
Solution:

Question 16.
The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms C.
Solution:
Assuming L along x-axis and C along y-axis, we have two points (124.942, 20) and (125.134, 110). By two point form, the point (L, C) satisfies the equation

Question 17.
The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs. 14/litre and 1220 litres of milk each week at Rs. 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17/litre?
Solution:
Assuming L (litres) along x-axis and R(rupees) along y-axis, we have two points (980,14) and (1220,16).
By two point form, the point (L, R) satisfies the equation.

Question 18.
P(a, b) is the mid-point of a lone segment between axes. Show that equation of the line is .
Solution:
Let the line AB makes intercepts c and d on the x-axis and y-axis respectively.

Question 19.
Point R(h, k) divides a line segment between the axes in the ratio 1:2. Find equation of the line.
Solution:
Let AB be the given line segment making intercepts a and b on the x-axis & y-axis respectively.
Then, the equation of line AB is

So, these points are A(a, 0) and B(0, b).
Now, R(h,k) divides the line segment Ab in the ratio 1 : 2.

Question 20.
By using the concept of equation of a line, prove that the three points (3, 0), (- 2, – 2) and (8, 2) are collinear.
Solution:
Let the given points be A(3, 0), B(-2, -2) and C(8, 2). Then the equation of the line passing through A and B is

Clearly the point C(8, 2) satisfy the equation 2x – 5y – 6 = 0.
(∵ 2(8) – 5(2) – 6 = 16 – 10 – 6 = 0)
Hence, the given points lie on the same straight line whose equation is 2x – 5y – 6 = 0.

Chapter 10 Straight Lines Exercise 10.3

Question 1.
Reduce the following equations into slope- intercept form and find their slopes and the y-intercepts.
(i) x + 7y = 0,
(ii) 6x + 3y-5 = 0,
(iii) y=0
Solution:
(i) We have given an equation x + 7y = 0, which can be written in the form
⇒ 7y = – x ⇒ y = x + 0 … (1)
Also, the slope intercept form is y-mx + c …(2)
On comparing (1) and (2), we get
m = , c = 0
Hence the slope is and the y-intercept = 0.

(ii) We have given an equation 6x + 3y – 5 = 0, which can be written in the form 3y = – 6x + 5
⇒ y = – 2x + …(1)
Also, the slope intercept form is y = mx + c … (2)
On comparing (1) and (2), we get
m = – 2 and c =
i.e. slope = – 2 and the y-intercept =

(iii) We have given an equation y = 0
y = 0·x + 0 … (1)
Also, the slope intercept form is y = mx + c … (2) On comparing (1) and (2), we get
m = 0, c = 0.
Hence, slope is 0 and the y-intercept is 0.

Question 2.
Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0,
(ii) 4x – 3y = 6,
(iii) 3y + 2 = 0
Solution:
(i) Given equation is 3x + 2y – 12 = 0 We have to reduce the given equation into intercept form, i.e., …(1)
Now given, 3x + 2y = 12
⇒ ⇒ …(2)
On comparing (1) and (2), we get a = 4, b = 6 Hence, the intercepts of the line are 4 and 6.

(ii) Given equation is 4x – 3y = 6
We have to reduce the given equation into intercept form, i.e., …(1)
…(2)
On comparing (1) and (2), we get
a = and b = – 2
Hence, the intercepts of the line are and -2.

(iii) Given equation is 3y + 2 = 0
We have to reduce the given equation into intercept form, i.e.,
3y = -2
⇒ y =
The above equation shows that, it is not the required equation of the intercept form as it is parallel to x-axis.
We observe that y-intercept of the line is , but there is no intercept on x-axis.

Question 3.
Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.
(i) x – + 8 = 0,
(ii) y-2 = 0,
(iii) x-y = 4.
Solution:
(i) Given equation is x – + 8 = 0
x – = -8
-x + = 8 … (i)
Also,

Now dividing both the sides of (1) by 2, we get

⇒ – cos 60°x + sin 60° y = 4.
⇒ {cos (180° – 60°)) x + {sin (180° – 60°)|y = 4
⇒ cos 120° x + sin 120° y = 4
∴ x cos 120° + y sin 120° = 4 is the required equation in normal form
∵ The normal form is x coso) + y sin⍵ = p
So, ⍵ = 120° and p = 4
⍵ Distance of the line from origin is 4 and the angle between perpendicular and positive x-axis is 120°.

(ii) Given equation is y – 2 = 0
⇒ y = 2
⇒ 0 · x + l · y = 2
⇒ x cos 90° + y sin 90° = 2 is the required equation in normal form
∵ The normal form is x cos⍵ + y sin⍵ = p
So, ⍵ = 90° and p = 2
⍵ Distance of the line from origin is 2 and the angle between perpendicular and positive x-axis is 90°.

(iii) Given equation is x – y = 4 … (1)
Also

Now dividing both the sides of (1) bt , we get

is the required equation in normal form.
∵ The normal form is x cos⍵ + y sin⍵ = p
So, P = and ⍵ = 315°
∴ Distance of the line from the origin is and the angle between perpendicular and the positive x-axis is 315°.

Question 4.
Find the distance of the point (-1, 1) from the line 12 (x+ 6) = 5(y — 2).
Solution:
The equation of line is 12(x + 6) = 5(y – 2) …(i)
⇒ 12x + 72 = 5y-10
⇒ 12x – 5y + 82 = 0
∴ Distance of the point (-1, 1) from the line (i)

Question 5.
Find the points on the x-axis, whose distances from the line are 4 units.
Solution:
We have a equation of line , which can be written as
4x + 3y – 12 = 0 … (i)
Let (a, 0) be the point on x-axis whose distance from line (i) is 4 units.

Question 6.
Find the distance between parallel lines
(i) 15x+8y-34 = 0and 15x + 8y+31 =0
(ii) |(x + y) + p = 0 and |(x + y) – r = 0.
Solution:
If lines are Ax + By + Q = 0
and Ax + By + C₂ = 0, then distance between

Question 7.
Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (-2, 3).
Solution:
We have given an equation of line 3x – 4y + 2 = 0
Slop of the line(i) =
Thus, slope of any line parallel to the given line (i) is and passes through (-2, 3), then its equation is

Question 8.
Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.
Solution:
Given equation is x – 7y + 5 = 0 … (i)
Slope of this line =
∴ Slope of any line perpendicular to the line (i) is -7 and passes through (3, 0) then
y – 0 = -7(x – 3)
[∵ Product of slope of perpendicular lines is -1]
⇒ y = -7x + 21
⇒ 7x + y – 21 = 0, is the required equation of line.

Question 9.
Find angles between the lines + y = 1 and x + = 1.
Solution:
The given equations are
+ y = 1 … (i)
x + = 1 … (ii)
Since we have to find an angle between the two lines i.e., firstly we have to find the slopes of (i) and (ii).

Question 10.
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.
Solution:
Given points are (h, 3) and (4,1).
∴ Slope of the line joining (h, 3) & (4,1)

Question 11.
Prove that the line through the point (x₁ y₁) and parallel to the line Ax + By + C = 0 is A(x-x₁) + B(y-y₁) = 0.
Solution:
Given equation of a line is Ax + By + C = 0
∴ Slope of the above line =
i.e. slope of any line parallel to given line and passing through (x₁, y₁) is
Then equation is (y – y₂) = (x – x₁)
=> B(y – y₁) = -A(x – x₁)
=> A(x – x₁) + B(y – y₁) = 0.
Hence proved.

Question 12.
Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.
Solution:
We have given a point (2, 3), through which two lines are passing and intersects at an angle of 60°.
Let m be the slope of the other line

Question 13.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
Solution:
suppose the given points are A and B.
Let M be the mid point of AB.

Question 14.
Find the coordinates of the foot of perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.
Solution:
We have, 3x – 4y – 16 = 0
Slope of the kine(i) =
Then equation of any line ⊥ from (-1, 3) to the given line(i) is

Question 15.
The perpendicular from the origin to the line y = mx + c meets it at the point (-1,2). Find the values of m and c.
Solution:
Given, the perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2)
∴ 2 = m (-1) + c … (i)
⇒ c – m = 2

Question 16.
If p and q are the lengths of perpendiculars from the origin to the lines x cosθ – y sinθ = k cos 2θ and x secθ + y cosecθ = k, respectively, prove that p² + 4q² = k².
Solution:
Given p and q are the lengths of perpendiculars from the origin to the lines x cos θ – ysinθ=k cos 2θ and xsecθ+y cosec θ = k.

Question 17.
In the triangle ABC with vertices A(2, 3), 8(4, -1) and C( 1, 2), find the equation and length of altitude from the vertex A.
Solution:
We have given a AABC with the vertices, A (2, 3), B (4, -1) and C (1, 2)

Question 18.
If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that .
Solution:
Given, p be the length of perpendicular from the origin to the line whose intercepts

We hope the NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines, help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines, drop a comment below and we will get back to you at the earliest.