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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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.
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
$=\frac{121}{2} \ units$
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 $( \frac{15}{2},0)$
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 $\frac{-1}{2}$
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
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 $-\sqrt3$
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 $135^\circ$
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
Also read
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
Students can refer subject wise NCERT solutions. The links to solutions are given below
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)
The angle made by line l with the positive direction of x-axis and measured anticlockwise is called the inclination of the line 'l'.
The tan θ is called the slope or gradient of line 'l' where θ is the inclination of line 'l'.
Slope of line = tan 45 = 1
Condition for collinearity of points A, B, and C.
Slope of AB = Slope of BC
The slope of the x-axis is zero.
The slope of the y-axis is not defined.
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