Exercise 6.2-Additional Questions - Chapter 6 Two Dimensional Analytical Geometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions Solved

Question 1.
Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of $120^{\circ}$ with the positive direction of $x$ - axis.
Solution:
Given that: $\quad \mathrm{OM}=4$ units

$
\begin{aligned}
\angle \mathrm{BAX} & =120^{\circ} \\
\therefore \angle \mathrm{BAO} & =180^{\circ}-120^{\circ} \text { or } \angle \mathrm{MAO}=60^{\circ} \\
\angle \mathrm{MOA}+\angle \mathrm{MAO} & =90^{\circ} \quad[\because \mathrm{OM} \perp \mathrm{AB}] \\
\theta+60^{\circ} & =90^{\circ} \quad \therefore \theta=30^{\circ} \quad
\end{aligned}
$
So, equation of $\mathrm{AB}$ in its normal form

$
\begin{gathered}
x \cos \theta+y \sin \theta=p \\
\Rightarrow \quad x \cos 30^{\circ}+y \sin 30^{\circ}=4 \\
\Rightarrow x \times \frac{\sqrt{3}}{2}+y \times \frac{1}{2}=4 \Rightarrow \sqrt{3} x+y=8
\end{gathered}
$
Hence, the required equation is $\sqrt{3} x+y=8$

Question 2.
Find the equation of the line which passes through the point $(-4,3)$ and the portion of the line intercepted between the axes is divided internally in the ratio $5: 3$ by this point.
Solution:

Let $\mathrm{AB}$ be a line passing through a point $(-4,3)$ and meets $\mathrm{x}$-axis at $\mathrm{A}(\mathrm{a}, 0)$ and $\mathrm{y}$-axis at $\mathrm{B}(0, \mathrm{~b})$.

$\left[\begin{array}{rr}
\because \mathrm{X} & =\frac{m_1 x_2+m_2 x_1}{m_1+m_2} \\
\text { and } \mathrm{Y} & =\frac{m_1 y_2+m_2 y_1}{m_1+m_2}
\end{array}\right]$

$\begin{aligned}
& & -4 & =\frac{5 \times 0+3 a}{5+3} \Rightarrow-4=\frac{3 a}{8} \\
\Rightarrow & & 3 a & =-32 \Rightarrow a=\frac{-32}{3} \\
& \text { and } & 3 & =\frac{5 . b+3.0}{5+3} \Rightarrow 3=\frac{5 b}{8} \\
\Rightarrow & & 5 b & =24 \Rightarrow b=\frac{24}{5}
\end{aligned}$

Intercept form of line is $\frac{x}{\frac{-32}{3}}+\frac{y}{\frac{24}{5}}=1 \Rightarrow \frac{-3 x}{32}+\frac{5 y}{24}=1$
$
\Rightarrow-9 x+20 y=96 \Rightarrow 9 x-20 y+96=0
$
Hence, the required equation is $9 x-20 y+96=0$

Question 3.
If the intercept of a line between the coordinate axes is divided by the point $(-5,4)$ in the ratio $1: 2$, then find the equation of the line.
Solution:
Let $\mathrm{a}$ and $\mathrm{b}$ be the intercepts on the given line.
Coordinates of $\mathrm{A}$ and $\mathrm{B}$ are $(a, 0)$ and $(0, b)$ respectively.

$\begin{aligned}
& -5=\frac{1 \times 0+2 \times a}{1+2} \Rightarrow 2 a=-15 \\
& \Rightarrow \quad a=\frac{-15}{2} \\
& \therefore \quad A=\left(\frac{-15}{2}, 0\right) \\
& \text { and } 4=\frac{1 \times b+0 \times 2}{1+2} \Rightarrow 4=\frac{b}{3} \Rightarrow b=12 \\
&
\end{aligned}$

$\left[\begin{array}{r}
\because \mathrm{X}=\frac{m_1 x_2+m_2 x_1}{m_1+m_2} \\
\text { and } \mathrm{Y}=\frac{m_1 y_2+m_2 y_1}{m_1+m_2}
\end{array}\right.$

$
\therefore \quad B=(0,12)
$
So the equation of line $\mathrm{AB}$ is $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

$
\begin{aligned}
& y-0=\left(\frac{12-0}{0+\frac{15}{2}}\right)\left(x+\frac{15}{2}\right) \\
& \Rightarrow \quad y=\frac{12 \times 2}{15}\left(x+\frac{15}{2}\right) \\
& \Rightarrow \quad y=\frac{8}{5}\left(x+\frac{15}{2}\right) \\
& \Rightarrow \quad 5 y=8 x+60 \Rightarrow 8 x-5 y+60=0
\end{aligned}
$
Hence, the required equation is $8 x-5 y+60=0$

Question 4.
Find the equation of the straight line which passes through the point $(1,-2)$ and cuts off equal intercepts from axes.
Solution:
Intercept form of a straight line is $\frac{x}{a}+\frac{y}{b}=1$, where $\mathrm{a}$ and $\mathrm{b}$ are the intercepts on the axis
Given that $a=b, \therefore \frac{x}{a}+\frac{y}{b}=1$
If equation (1) passes through the point $(1,-2)$ we get $\frac{1}{a}-\frac{2}{a}=1 \Rightarrow-\frac{1}{a}=1 \Rightarrow a=-1$
So, equation of the straight line is $\mathrm{x} \mathrm{v}$
$
\frac{x}{-1}+\frac{y}{-1}=1 \Rightarrow x+y=-1 \Rightarrow x+y+1=0
$
Hence, the required equation $x+y+1=0$
Question 5.
Find the distance of the line $4 x-y=0$ from the point $\mathrm{P}(4,1)$ measured along the line making an angle $135^{\circ}$ with the positive $\mathrm{x}$-axis
Solution:
The equation in distance form of the line passing through $\mathrm{P}(4,1)$ and making an angle of $135^{\circ}$ with the positive $\mathrm{x}$-axis

$
\frac{x-4}{\cos 135^{\circ}}=\frac{y-1}{\sin 135^{\circ}}
$
Suppose it cuts $4 x-y=0$ at $\mathrm{Q}$ such that $\mathrm{PQ}=r$ Then, the coordinates of $\mathrm{Q}$ are given by
$
\frac{x-4}{\cos 135^{\circ}}=\frac{y-1}{\sin 135^{\circ}}=r
$

$
\begin{aligned}
& \Rightarrow \frac{x-4}{-1 / \sqrt{2}}=\frac{y-1}{1 / \sqrt{2}}=r \\
& \Rightarrow x=4-\frac{r}{\sqrt{2}}, y=1+\frac{r}{\sqrt{2}}
\end{aligned}
$
So, the coordinates of $Q$ are $\left(4-\frac{r}{\sqrt{2}}, 1+\frac{r}{\sqrt{2}}\right)$
Clearly, Q lies on $4 x-y=0$
$
\begin{aligned}
& 4\left(4-\frac{r}{\sqrt{2}}\right)-\left(1+\frac{r}{\sqrt{2}}\right)=0 \\
& \therefore 16-\frac{4 r}{\sqrt{2}}-1-\frac{r}{\sqrt{2}}=0 \Rightarrow \frac{5 r}{\sqrt{2}}=15 \Rightarrow r=3 \sqrt{2}
\end{aligned}
$
Hence, required distance is $3 \sqrt{2}$ units. .
Question 6.
The line $2 \mathrm{x}-\mathrm{y}=5$ turns about the point on it, whose ordinate and abscissa are equal, through an angle of $45^{\circ}$ in the anti-clockwise direction, find the equation of the line in the new position.
Solution:
If the line $2 \mathrm{x}-\mathrm{y}=5$ makes an angle $\theta$ with $x$ - axis. Then, $\tan \theta=2$. Let $P(\alpha, \alpha)$ be a point on the line $2 \mathrm{x}-\mathrm{y}=5$.
Then, $2 \alpha-\alpha=5 \Rightarrow \alpha=5$

So, the coordinates of $P$ are $(5,5)$. If the line $2 x-y-5=0$ is rotated about point $\mathrm{P}$ through $45^{\circ}$ in anti clockwise direction, then the line in its new position makes angle $\theta+45^{\circ}$ with $\mathrm{x}-$ axis. Let $m$ ' be the slope of the line in its new position. Then,
$
m^{\prime}=\tan \left(\theta+45^{\circ}\right)=\frac{\tan \theta+\tan 45^{\circ}}{1-\tan \theta \tan 45^{\circ}}=\frac{2+1}{1-2 \times 1}=-3
$
Thus, the line in its new pdsition passes through $P(5,5)$ and has slope $m^{\prime}=-3$
So, its equation $y-5=m^{\prime}(x-5)$ or, $y-5=-3(x-5)$ or, $3 x+y-20=0$

Let $y=m_1 x+c_1$ and $y=m_2 x+c_2$ be the equations of two straight lines and let these two lines make angles $\theta_1$ and $\theta_2$ with $x$ - axis.
Then $m_1=\tan \theta_1$ and $m_2=\tan \theta_2$
If $\phi$ (phi) is the angle between these two straight lines, then $\Rightarrow \tan \phi=\frac{m_2-m_1}{1+m_2 m_1} \Rightarrow \phi=\tan ^{-1} \frac{m_2-m_1}{1+m_2 m_1}$