Exercise 6.2-Additional Questions - Chapter 6 - Two Dimensional Analytical Geometry - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 6.2-Additional Questions - Chapter 6 - Two Dimensional Analytical Geometry - 11th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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 $\Rightarrow \tan \phi=\frac{m_2-m_1}{1+m_2 m_1} \Rightarrow \phi=\tan ^{-1} \frac{m_2-m_1}{1+m_2 m_1}$