Exercise 9.1 (Revised) - Chapter 10 - Straight Line - Ncert Solutions class 11 - Maths

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Chapter 9: Straight Line - NCERT Solutions for Class 11 Maths | Detailed Explanations

Ex 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.

Let $A(-4,5), B(0,7), C(5,-5)$ and $D(-4,-2)$ be the vertices of the quadrilateral Join the vertices $\mathrm{A}$ and $\mathrm{C}$ to obtain the diagonal $\mathrm{AC}$

Area of quadrilateral $\mathrm{ABCD}=$ Area of triangle $\mathrm{ABC}+$ Area of triangle $\mathrm{ACD}$.
But we have area of a triangle with vertices $\left(x_1, y_1\right),\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ is
$
\begin{aligned}
& \frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\
& \therefore \text { Area of triangle ABC }=\frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)| \\
& =\frac{1}{2}|-4(12)+0+5(-2)|=\frac{1}{2}|-48+0-10|=\frac{58}{2} \text { square units }
\end{aligned}
$

Also Area of triangle ACD $=\frac{1}{2}|-4(-5+2)+5(-2-5)+(-4)(5+5)|$ $=\frac{1}{2}|-4(-3)+5(-7)+(-4)(10)|=\frac{1}{2}|12-35-40|=\frac{63}{2}$ square units
$\therefore$ From equation (i) we haveArea of quadrilateral $\mathrm{ABCD}=\frac{58}{2}+\frac{63}{2}=\frac{121}{2}$ square units

Ex 9.1 Question 2.

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

Answer.

Given: Length of side of equilateral triangle $=2 a$. The base of triangle lies along $y-$ axis and the mid-point of base is at origin so that the coordinates of vertices are $(0, a)$ and $(0,-a)$.

We have for an equilateral triangle line from the vertex to the mid point of the base will be perpendicular to the base.So we can say the third vertex lies on $\mathrm{X}$ axis (positive or negative)

Now let the third vertex be $( \pm x, 0)$.
It is known that area of an equilaterl triangle $=\frac{\sqrt{3}}{4} a^2$, where a is the common length of the sides
$\therefore$ Area of theequilaterl triangle with side $2 \mathrm{a}=\frac{\sqrt{3}}{4}(2 a)^2$

$
\begin{aligned}
& \Rightarrow \frac{1}{2} \times 2 a \times( \pm x)=\frac{\sqrt{3}}{4} \times(2 a)^2 \\
& \Rightarrow x= \pm \sqrt{3} a
\end{aligned}
$

Therefore the third vertex can be $(\sqrt{3} a \quad, 0)$ or $(-\sqrt{3} a, 0)$
$\therefore$ The vertices of triangle are $(0, a),(0,-a)$ and $(\sqrt{3} a, 0)$ or $(0, a),(0,-a)$ and $(-\sqrt{3} a, 0)$
Ex 9.1 Question 3.

Find the distance between $\mathrm{P}\left(x_1, y_1\right)$ and $\mathrm{Q}\left(x_2, y_2\right)$ when (i) $\mathrm{PQ}$ is parallel to the $y$ axis (ii) $P Q$ is parallel to the $x$-axis.

Answer.

Given: $\mathrm{P}\left(x_1, y_1\right)$ and $\mathrm{Q}\left(x_2, y_2\right)$ are two points.
$
\therefore \mathrm{PQ}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}
$
(i) We have equation of a line parallel to $y$-axis is $X=K$, a constant $\mathrm{PQ}$ is parallel to $y$-axis, then $x_2=x_1 \Rightarrow x_2-x_1=0$
$
\therefore \mathrm{PQ}=\sqrt{\left(y_2-y_1\right)^2}=\left|y_2-y_1\right|
$
(ii) We have equation of a line parallel to $x$-axis is $\mathrm{Y}=\mathrm{K}$, a constant

$\mathrm{PQ}$ is parallel to $x$-axis, then $y_2-y_1=0$.
$
\therefore \mathrm{PQ}=\sqrt{\left(x_2-x_1\right)^2}=\left|x_2-x_1\right|
$
Ex 9.1 Question 4.

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

Answer.

Let $\mathrm{P}(x, 0)$ be any point on the $x$-axis which is equidistant from $\mathrm{Q}(7,6)$ and $\mathrm{R}(3,4)$.
We have distnce between two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$
\begin{aligned}
& \therefore P Q=\sqrt{(x-7)^2+(0-6)^2} \\
& =\sqrt{x^2-14 x+49+36} \\
& =\sqrt{x^2-14 x+85}
\end{aligned}
$

And PR $=\sqrt{(x-3)^2+(0-4)^2}$
$
\begin{aligned}
& =\sqrt{x^2-6 x+9+16} \\
& =\sqrt{x^2-6 x+25}
\end{aligned}
$

According to question, $\mathrm{PQ}=\mathrm{PR}$
$
\Rightarrow \sqrt{x^2-14 x+85}=\sqrt{x^2-6 x+25}
$

Squaring both sides, $x^2-14 x+85=x^2-6 x+25$
$
\begin{aligned}
& \Rightarrow-14 x+6 x=25-85 \\
& \Rightarrow-8 x=-60 \Rightarrow x=\frac{15}{2}
\end{aligned}
$

Therefore, the required point is $\left(\frac{15}{2}, 0\right)$.

Ex 9.1 Question 5.

Find the slope of a line, which passes through the origin and the mid-point of the line segment of joining the points $P(0,-4)$ and $B(8,0)$.

Answer.

Here, mid-point of the line segment joining $P(0,-4)$ and $Q(8,0)$ is $\left(\frac{0+8}{2}, \frac{-4+0}{2}\right)=(4=-2)$ using mid point formula

We have slope of a line passing through two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is given by

$
m=\frac{y_2-y_1}{x_2-x_1}
$

Therefore, Slope of the line passing through the points $(0,0)$ and $(4,-2) .=$
$
\frac{-2-0}{4-0}=\frac{-2}{4}=\frac{-1}{2}
$

Hence slope of the required line is $\frac{-1}{2}$
Ex 9.1 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.

Let $\mathrm{A}(4,4), \mathrm{B}(3,5)$ and $\mathrm{C}(-1,-1)$ be the three vertices of $\triangle \mathrm{ABC}$
We have slope of a line passing through two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$
$\therefore$ Slope of $\mathrm{AB}=\frac{5-4}{3-4}=\frac{1}{-1}=-1$
Slope of $\mathrm{BC}=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}$
Slope of $\mathrm{AC}=\frac{-1-4}{-1-4}=\frac{-5}{-5}=1$

Now, Slope of $\mathrm{AB} \times$ Slope of $\mathrm{AC}=-1 \times 1=-1$
This shows that $\mathrm{AB} \perp \mathrm{AC}$. Thus $\triangle \mathrm{ABC}$ is right angled at point $\mathrm{A}$.
Ex 9.1 Question 7.

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

If the line makes an angle of $30^{\circ}$ with the positive direction of $y$-axis then the line will make an angle of $\left(90^{\circ}+30^{\circ}\right)=120^{\circ}$ with the positive direction of $x$-axis.
$\therefore$ Slope of the line $=\tan 120^{\circ}=\tan \left(90^{\circ}+30^{\circ}\right)=-\cot 30^{\circ}=-\sqrt{3}$

Ex 9.1 Question 8

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

Answer.

Let $\mathrm{A}(-2,-1), \mathrm{B}(4,0), \mathrm{C}(3,3)$ and $\mathrm{D}(-3,2)$ be vertices of a quadrilateral $\mathrm{ABCD}$.
To prove a quadrilateral is a parallelogram it is enough to show that both pairs of opposite sides are parallel

We have slope of a line passing through two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$
$\therefore$ Slope of $\mathrm{AB}=\frac{0-(-1)}{4-(-2)}=\frac{1}{6}$ Slope of $\mathrm{BC}=\frac{3-0}{3-4}=\frac{3}{-1}=-3$
Slope of DC $=\frac{3-2}{3-(-3)}=\frac{1}{6}$ Slope of $\mathrm{AD}=\frac{2-(-1)}{-3-(-2)}=\frac{3}{-1}=-3$
Here Slope of $\mathrm{AB}=$ Slope of $\mathrm{DC}$
$\Rightarrow \mathrm{AB} \| \mathrm{DC}$
And Slope of $\mathrm{BC}=$ Slope of $\mathrm{AD}$
$
\Rightarrow \mathrm{BC} \| \mathrm{AD}
$

Therefore, ABCD is a parallelogram.

Ex 9.1 Question 9

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

Answer.

Let $\mathrm{A}(3,-1)$ and $\mathrm{B}(4,-2)$ be two points. Let $\theta$ be the angle which $\mathrm{AB}$ makes with positive direction of $x$ - axis.
$\therefore$ Slope of $\mathrm{AB}=\tan \theta$
We have slope of a line passing through two points $\left(x_1, y_1\right)$ and $\left(x_2=y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$

$\therefore$ Slope of $A B=\frac{-2-(-1)}{4-3}=\frac{-1}{1}=-1$
From (i) and (ii) we get $\tan \theta=-1$
$
\begin{aligned}
& \Rightarrow \tan \theta=-\tan 45^{\circ} \\
& \Rightarrow \tan \theta=\tan \left(180^{\circ}-45^{\circ}\right) \\
& \Rightarrow \tan \theta=\tan 135^{\circ} \\
& \Rightarrow \theta=135^{\circ}
\end{aligned}
$

Ex 9.1 Question 10.

The slope of a line is double of the slope of the another line. If tangent of the angle between them is $1 / 3$ find the slopes of the lines.

Answer.

Given: $\tan \theta=\frac{1}{3}$. Let the slopes of two lines be $m$ and $2 m$.
We know that if $\theta$ is the acute angle between two lines with slopes $m_1$ and $m_2$ respectively then $\tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|$
$\therefore\left|\frac{m-2 m}{1+m \times 2 m}\right|=\frac{1}{3}$
$\Rightarrow\left|\frac{-m}{1+2 m^2}\right|=\frac{1}{3}$
$\Rightarrow \frac{-m}{1+2 m^2}= \pm \frac{1}{3}$
Taking $\frac{-m}{1+2 m^2}=\frac{1}{3}$
$\Rightarrow-3 m=1+2 m^2$
$\Rightarrow 2 m^2+3 m+1=0$

$
\begin{aligned}
& \Rightarrow(m+1)(2 m+1)=0 \\
& \Rightarrow m=-1 \text { and } m=\frac{-1}{2}
\end{aligned}
$

When $m=-1$ the slopes of the lines are -1 and -2 when $m=\frac{-1}{2}$ the slopes of the lines are $\frac{-1}{2}$ and -1
$
\begin{aligned}
& \text { Taking } \frac{-m}{1+2 m^2}=-\frac{1}{3} \\
& \Rightarrow-3 m=-1-2 m^2 \\
& \Rightarrow 2 m^2-3 m+1=0 \\
& \Rightarrow(m-1)(2 m-1)=0 \\
& \Rightarrow m=1 \text { and } m=\frac{1}{2}
\end{aligned}
$

When $\mathrm{m}=1$ the slopes of the lines are 1 and 2 when $m=\frac{1}{2}$ the slopes of the lines are $\frac{1}{2}$ and 1

Therefore, the slopes of lines are -1 and -2 or 1 and 2 or -1 and $\frac{-1}{2}$ or 1 and $\frac{1}{2}$.
Ex 9.1 Question 11.

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

Answer.

Let $\mathrm{A}\left(x_1, y_1\right)$ and $\mathrm{B}(h, k)$ be two points. It is given that Slope of $\mathrm{AB}=m$

We have slope of a line passing through two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$

$\therefore$ Slope of $\mathrm{AB}=\frac{k-y_1}{h-x_1}=m$ (given)
$
\Rightarrow k-y_1=m\left(h-x_1\right)
$

Hence proved