NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1 - Straight Lines

NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1 - Straight Lines

Komal MiglaniUpdated on 06 May 2025, 02:49 PM IST

A straight line has various real-world applications, such as in architecture, engineering, art, navigation, and even in video games. In the Class 11 maths exercise 9.1, you will learn about the basic coordinate geometry, like measuring the slope of the line, conditions for parallelism and perpendicularity of lines, angle between two lines, colinearity of three points, etc. Understanding straight lines helps you analyze motion and direction, interpret graphs and data visually, and design and build real-world structures. Also, check NCERT to learn more about a straight line.

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  1. Class 11 Maths Chapter 9 Exercise 9.1 Solutions - Download PDF
  2. NCERT Solutions for Class 11 Maths Chapter 9 – Straight Lines Exercise 9.1
  3. Topics covered in Chapter 9 straight lines Exercise 9.1
  4. NCERT Solutions of Class 11 Subject Wise
  5. Subject-Wise NCERT Exemplar Solutions
NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.1 - Straight Lines
10.1

Our experienced subject experts designed the NCERT solutions straight line exercise 9.1 to offer a systematic and structured approach to these important concepts and help students to develop a clear understanding of critical concepts by a series of solved questions and conceptual explanations. These NCERT Solutions follow the CBSE pattern and also help in gaining knowledge about all the natural processes happening around them. These solutions provide a valuable resource to the students to enhance their performance in board exams.


Class 11 Maths Chapter 9 Exercise 9.1 Solutions - Download PDF

Download PDF


NCERT Solutions for Class 11 Maths Chapter 9 – Straight Lines Exercise 9.1

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.

Answer:



Area of ABCD = Area of ABC + Area of ACD
Now, we know that the area of a triangle with vertices $(x_1,y_1),(x_2,y_2) \ and \ (x_3,y_3)$ is given by
$A = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
Therefore,
Area of triangle ABC $= \frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)|$

$= \frac{1}{2}|-48-10|= \frac{58}{2}=29$

Similarly,
Area of triangle ACD $= \frac{1}{2}|-4(-5+2)+5(-2-5)+-4(5+5)|$

$= \frac{1}{2}|12-35-40|= \frac{63}{2}$
Now,
Area of ABCD = Area of ABC + Area of ACD
$=\frac{121}{2} \ units$

Question 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 vertices of the triangle.

Answer:


it is given that it is an equilateral triangle and length of all sides is 2a
The base of the triangle lies on y-axis such origin is the midpoint
Therefore,
Coordinates of point A and B are $(0,a) \ \ and \ \ (0,-a)$ respectively
Now,
Apply Pythagoras theorem in triangle AOC
$AC^2=OA^2+OC^2$
$(2a)^2=a^2+OC^2$
$OC^2= 4a^2-a^2=3a^2$
$OC=\pm \sqrt3 a$
Therefore, coordinates of vertices of the triangle are
$(0,a),(0,-a) \& \ (\sqrt3a,0) \ \ or \ \ (0,a),(0,-a) \& \ (-\sqrt3a,0)$

Question 3:(i) Find the distance between $P(x_1,y_1)$ and $Q(x_2,y_2)$ when :

PQ is parallel to the $y$-axis.

Answer:

When PQ is parallel to the y-axis
then, x coordinates are equal i.e. $x_2 = x_1$
Now, we know that the distance between two points is given by
$D = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|$
Now, in this case $x_2 = x_1$
Therefore,
$D = |\sqrt{(x_2-x_2)^2+(y_2-y_1)^2}| $

$= |\sqrt{(y_2-y_1)^2}|= |(y_2-y_1)|$
Therefore, the distance between $P(x_1,y_1)$ and $Q(x_2,y_2)$ when PQ is parallel to y-axis is $|(y_2-y_1)|$

Question 3:(ii) Find the distance between $P(x_1,y_1)$ and $Q(x_2,y_2)$ when :

PQ is parallel to the $x$-axis.

Answer:

When PQ is parallel to the x-axis
then, x coordinates are equal i.e. $y_2 = y_1$
Now, we know that the distance between two points is given by
$D = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|$
Now, in this case $y_2 = y_1$
Therefore,
$D = |\sqrt{(x_2-x_1)^2+(y_2-y_2)^2}| $

$= |\sqrt{(x_2-x_1)^2}|= |x_2-x_1|$
Therefore, the distance between $P(x_1,y_1)$ and $Q(x_2,y_2)$ when PQ is parallel to the x-axis is $|x_2-x_1|$

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

Answer:

Point is on the x-axis, therefore, y coordinate is 0
Let's assume the point is (x, 0)
Now, it is given that the given point (x, 0) is equidistance from point (7, 6) and (3, 4)
We know that
Distance between two points is given by
$D = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|$
Now,
$D_1 = |\sqrt{(x-7)^2+(0-6)^2}|= |\sqrt{x^2+49-14x+36}|$

$= |\sqrt{x^2-14x+85}|$
and
$D_2 = |\sqrt{(x-3)^2+(0-4)^2}|= |\sqrt{x^2+9-6x+16}|$

$= |\sqrt{x^2-6x+25}|$
Now, according to the given condition
$D_1=D_2$
$|\sqrt{x^2-14x+85}|= |\sqrt{x^2-6x+25}|$
Squaring both the sides
$x^2-14x+85= x^2-6x+25\\ 8x = 60\\ x=\frac{60}{8}= \frac{15}{2}$
Therefore, the point is $( \frac{15}{2},0)$

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 $B(8,0)$.

Answer:

Mid-point of the line joining the points $P(0,-4)$ and $B(8,0)$. is
$l = \left ( \frac{8}{2},\frac{-4}{2} \right ) = (4,-2)$
It is given that line also passes through origin which means passes through the point (0, 0)
Now, we have two points on the line so we can now find the slope of a line by using formula
$m = \frac{y_2-y_1}{x_2-x_1}$
$m = \frac{-2-0}{4-0} = \frac{-2}{4}= \frac{-1}{2}$
Therefore, the slope of the line is $\frac{-1}{2}$

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.

Answer:

It is given that point A(4,4) , B(3,5) and C(-1,-1) are the vertices of a right-angled triangle
Now,
We know that the distance between two points is given by
$D = |\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}|$
Length of AB $= |\sqrt{(4-3)^2+(4-5)^2}|= |\sqrt{1+1}|= \sqrt2$
Length of BC $= |\sqrt{(3+1)^2+(5+1)^2}|= |\sqrt{16+36}|= \sqrt{52}$
Length of AC $= |\sqrt{(4+1)^2+(4+1)^2}|= |\sqrt{25+25}|= \sqrt{50}$
Now, we know that Pythagoras theorem is
$H^2= B^2+L^2$
Is clear that
$(\sqrt{52})^2=(\sqrt{50})^2+(\sqrt 2)^2\\$

$52 = 52$

$\\ i.e\\ BC^2= AB^2+AC^2$
Hence proved

Question 7: Find the slope of the line, which makes an angle of $30^{\circ}$ with the positive direction of $y$-axis measured anticlockwise.

Answer:

It is given that the line makes an angle of $30^{\circ}$ with the positive direction of $y$-axis measured anticlockwise
Now, we know that
$m = \tan \theta$
line makes an angle of $30^{\circ}$ with the positive direction of $y$-axis
Therefore, the angle made by line with the positive x-axis is = $90^{^\circ}+30^{^\circ}= 120^\circ$
Now,
$m = \tan 120^\circ = -\tan 60^\circ = -\sqrt3$
Therefore, the slope of the line is $-\sqrt3$

Question 8: Without using the distance formula, show that points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are the vertices of a parallelogram.

Answer:

Given points are $A(-2,-1),B(4,0),C(3,3)$ and $D(-3,2)$
We know the pair of the opposite side are parallel to each other in a parallelogram
Which means their slopes are also equal
$Slope = m = \frac{y_2-y_1}{x_2-x_1}$
The slope of AB =

$\frac{0+1}{4+2} = \frac{1}{6}$

The slope of BC =

$\frac{3-0}{3-4} = \frac{3}{-1} = -3$

The slope of CD =

$\frac{2-3}{-3-3} = \frac{-1}{-6} = \frac{1}{6}$

The slope of AD

= $\frac{2+1}{-3+2} = \frac{3}{-1} = -3$
We can clearly see that
The slope of AB = Slope of CD (which means they are parallel)
and
The slope of BC = Slope of AD (which means they are parallel)
Hence pair of opposite sides are parallel to each other
Therefore, we can say that points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are the vertices of a parallelogram

Question 9: Find the angle between the x-axis and the line joining the points $(3,-1)$ and $(4,-2)$.

Answer:

We know that
$m = \tan \theta$
So, we need to find the slope of line joining points (3,-1) and (4,-2)
Now,
$m = \frac{y_2-y_1}{x_2-x_1}= \frac{-2+1}{4-3} = -1$
$\tan \theta = -1$
$\tan \theta = \tan \frac{3\pi}{4} = \tan 135^\circ$
$\theta = \frac{3\pi}{4} = 135^\circ$
Therefore, angle made by line with positive x-axis when measure in anti-clockwise direction is $135^\circ$

Question 10: The slope of a line is double of the slope of another line. If tangent of the angle between them is $\frac{1}{3},$ find the slopes of the lines

Answer:

Let $m_1 \ and \ m_2$ are the slopes of lines and $\theta$ is the angle between them
Then, we know that
$\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |$
It is given that $m_2 = 2m_1$ and

$\tan \theta = \frac{1}{3}$
Now,
$\frac{1}{3}= \left | \frac{2m_1-m_1}{1+m_1.2m_1} \right |$
$\frac{1}{3}= \left | \frac{m_1}{1+2m^2_1} \right |$
Now,
$3|m_1|= 1+2|m^2_1|\\ 2|m^2_1|-3|m_1|+ 1 $

$= 0\\ 2|m^2_1|-2|m_1|-|m_1|+1$

=$0\\ (2|m_1|-1)(|m_1|-1)= 0\\ |m_1|$

$= \frac{1}{2} \ \ \ \ \ or \ \ \ \ \ \ |m_1| = 1$
Now,
$m_1 = \frac{1}{2} \ or \ \frac{-1}{2} \ or \ 1 \ or \ -1$
According to which value of $m_2 = 1 \ or \ -1 \ or \ 2 \ or \ -2$
Therefore, $m_1,m_2 = \frac{1}{2},1 \ or \ \frac{-1}{2},-1 \ or \ 1,2 \ or \ -1,-2$

Question 11: A line passes through $(x_1,y_1)$ and $(h,k)$ . If slope of the line is $m$, show that $k-y_1=m(h-x_1)$.

Answer:

Given that A line passes through $(x_1,y_1)$ and $(h,k)$ and slope of the line is m
Now,
$m = \frac{y_2-y_1}{x_2-x_1}$
$\Rightarrow m = \frac{k-y_1}{h-x_1}$
$\Rightarrow (k-y_1)= m(h-x_1)$
Hence proved

Topics covered in Chapter 9 straight lines Exercise 9.1

1) Slope of a Line:

If θ is the inclination of a line L, then tan θ is called the slope of the line L. The slope of a line whose inclination is 90° is not defined. It may be observed that the slope of the x-axis is zero, and the slope of the y-axis is not defined. The slope of a line is denoted by m.

2) Slope of a line when the coordinates of any two points on the line are given:

Let P(x1, y1) and Q(x2, y2), then the slope of the line which is passing through these points will be m:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

3)Conditions for parallelism and perpendicularity of lines in terms of their slope:

If the line L1 ( slope m1) is parallel to the line L2 (slope m2), then their slopes will always be equal; therefore, m1 = m2.

The lines L1 ( slope m1) and the line L2 (slope m2) are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

$m_{2}=-\frac{1}{m_{1}}$ or $m_{1}\times m_{2}=-1$


4)Angle between two lines :

Let L1 and L2 be two non-vertical lines with slopes m1 and m2, and theta is the angle between them, then theta will be

$tan\Theta =\left | \frac{m_{2}-m_{1}}{1+m_{1}m_{2}} \right |$

where $1+m_{1}m_{2}\neq 0$

Also Read

NCERT Solutions of Class 11 Subject Wise

Students can refer subject wise NCERT solutions. The links to solutions are given below


Subject-Wise NCERT Exemplar Solutions

Students can access the NCERT exemplar solutions to enhance their deep understanding of the topic. These solutions are aligned with the CBSE syllabus and also help in competitive exams.



Frequently Asked Questions (FAQs)

Q: What is the inclination of a line 'l' ?
A:

The angle made by line l with the positive direction of x-axis and measured anticlockwise is called the inclination of the line 'l'.

Q: What is the slope of a line ?
A:

The tan θ is called the slope or gradient of line 'l' where θ is the inclination of line 'l'.

Q: If the inclination of line in 45 degree than what is the slope of the line ?
A:

Slope of line = tan 45 = 1

Q: Let ABC are three points on xy-plane, than what is the condition for collinearity of ABC ?
A:

Condition for collinearity of points A, B, and C.

Slope of AB = Slope of BC

Q: What is the slope of x-axis ?
A:

The slope of the x-axis is zero.

Q: What is the slope of y-axis ?
A:

The slope of the y-axis is not defined.

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