NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.1

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.1 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.1.

• Straight Lines Class 11 Ex 10.2
• Straight Lines Class 11 Ex 10.3

Board	CBSE
Textbook	NCERT
Class	Class 11
Subject	Maths
Chapter	Chapter 10
Chapter Name	Straight Lines
Exercise	Ex 10.1
Number of Questions Solved	14
Category	NCERT Solutions

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 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

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