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

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
$A{C}^2=O{A}^2+O{C}^2$
$(2a{)}^2=a^2+O{C}^2$
$O{C}^2=4a^2-a^2=3a^2$
$OC=\pm \sqrt{3}a$
Therefore, coordinates of vertices of the triangle are
$(0,a),(0,-a)\mathrm{\&}(\sqrt{3}a,0)or(0,a),(0,-a)\mathrm{\&}(-\sqrt{3}a,0)$

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
$$\begin{gathered} x^2-14x+85=x^2-6x+25 \\ 8x=60 \\ x=\frac{60}{8}=\frac{15}{2} \end{gathered}$$
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
$l=(\frac{8}{2},\frac{-4}{2})=(4,-2)$
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 $=|\sqrt{(4-3{)}^2+(4-5{)}^2}|=|\sqrt{1+1}|=\sqrt{2}$
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
$$\begin{gathered} (\sqrt{52}{)}^2=(\sqrt{50}{)}^2+(\sqrt{2}{)}^2 \\ 52=52 \\ i.e \\ B{C}^2=A{B}^2+A{C}^2 \end{gathered}$$
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,
$m=\tan {120}^{\circ }=-\tan {60}^{\circ }=-\sqrt{3}$
Therefore, the slope of the line is $-\sqrt{3}$

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

Answer:

Point is collinear which means they lie on the same line by this we can say that their slopes are equal
Given points are A(x,-1) , B(2,1) and C(4,5)
$Slope=m=\frac{y_2-y_1}{x_2-x_1}$
Now,
The slope of AB = Slope of BC
$\frac{1+1}{2-x}=\frac{5-1}{4-2}$
$$\begin{gathered} \frac{2}{2-x}=\frac{4}{2} \\ \frac{2}{2-x}=2 \\ 2=2(2-x) \\ 2=4-2x \\ -2x=-2 \\ x=1 \end{gathered}$$
Therefore, the value of x is 1

Question:9 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:10 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:11 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
$\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$
It is given that $m_2=2m_1$ and

$\tan \theta =\frac{1}{3}$
Now,
$\frac{1}{3}=\left|\frac{2m_1-m_1}{1+m_1.2m_1}\right|$
$\frac{1}{3}=\left|\frac{m_1}{1+2m_1^2}\right|$
Now,
$$\begin{gathered} 3|m_1|=1+2|m_1^2| \\ 2|m_1^2|-3|m_1|+1=0 \\ 2|m_1^2|-2|m_1|-|m_1|+1=0 \\ (2|m_1|-1)(|m_1|-1)=0 \\ |m_1|=\frac{1}{2}or|m_1|=1 \end{gathered}$$
Now,
$m_1=\frac{1}{2}or\frac{-1}{2}or1or-1$
According to which value of $m_2=1or-1or2or-2$
Therefore, $m_1,m_2=\frac{1}{2},1or\frac{-1}{2},-1or1,2or-1,-2$

Question:12 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

Question:13 If three points $(h,0),(a,b)$ and $(0,k)$ lie on a line, show that $\frac{a}{h}+\frac{b}{k}=1.$

Answer:

Points $A(h,0),B(a,b)$ and $C(0,k)$ lie on a line so by this we can say that their slopes are also equal
We know that
$Slope=m=\frac{y_2-y_1}{x_2-x_1}$

Slope of AB = $\frac{b-0}{a-h}=\frac{b}{a-h}$

Slope of AC = $\frac{k-b}{0-a}=\frac{k-b}{-a}$
Now,
Slope of AB = slope of AC
$\frac{b}{a-h}=\frac{k-b}{-a}$
$-ab=(a-h)(k-b)$
$$\begin{gathered} -ab=ak-ab-hk+hb \\ ak+hb=hk \end{gathered}$$
Now divide both the sides by hk
$$\begin{gathered} \frac{ak}{hk}+\frac{hb}{hk}=\frac{hk}{hk} \\ \frac{a}{h}+\frac{b}{k}=1 \end{gathered}$$
Hence proved

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?

Answer:

Given point A(1985,92) and B(1995,97)
Now, we know that
$Slope=m=\frac{y_2-y_1}{x_2-x_1}$
$m=\frac{97-92}{1995-1985}=\frac{5}{10}=\frac{1}{2}$
Therefore, the slope of line AB is $\frac{1}{2}$
Now, the equation of the line passing through the point (1985,92) and with slope = $\frac{1}{2}$ is given by
$$\begin{gathered} (y-92)=\frac{1}{2}(x-1985) \\ 2y-184=x-1985 \\ x-2y=1801 \end{gathered}$$
Now, in the year 2010 the population is
$$\begin{gathered} 2010-2y=1801 \\ -2y=-209 \\ y=104.5 \end{gathered}$$
Therefore, the population in the year 2010 is 104.5 crore