The point $\mathrm{P}(2,1)$ is translated to a point $\mathrm{Q}$ parallel to the line $L \equiv x-y-4=0$ by $2 \sqrt{3}$ units. If the point $Q$ lies in the third quadrant, then the equation of the line passing through $\mathrm{Q}$ and perpendicular to $\mathrm{L}$ is

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Aicharya · Accepted Answer

Given $L \equiv x-y-4=0$ $ \Rightarrow \frac{x}{(-4)}+\frac{y}{(4)}=1 $ Slope $m_1$ of line ' $L$ ' is $m_1=1$

Since line ' $L$ ' and line $P Q$ is parallel to each other. Hence slope of line ' $L$ ' $=$ Slope of line $P Q$ Hence equation of $P Q$ can be written as $ \begin{aligned} & \Rightarrow y-1=m_1(x-2)=(1)(x-2) \\ & \Rightarrow y=(x-1) \\ & \because \text { Distance } P Q=2 \sqrt{3} \\ & \Rightarrow \sqrt{(x-2)^2+(y-1)^2}=2 \sqrt{3} \\ & \Rightarrow(x-2)^2+(x-2)^2=12 \\ & \Rightarrow(x-2)^2=6 \\ & \Rightarrow(x-2)= \pm \sqrt{6} \text { or } x=2+\sqrt{6} \text { or } 2-\sqrt{6} \end{aligned} $ Here $x=2+\sqrt{6}$ is not valid because point $Q$ lies in third quadrant. Hence $x=2-\sqrt{6}$ Therefore $y=(2-\sqrt{6})-1=1-\sqrt{6}$ $ \therefore Q(x, y)=Q(2-\sqrt{6}, 1-\sqrt{6}) $ Let the slope of line perpendicular to given line ' $L$ ' is $m_2$. Then, $ m_1 m_2=-1 \Rightarrow m_2=-\frac{1}{m_1}=-\frac{1}{1}=-1 $ Hence equation of line perpendicular to ' $L$ ' and passing through point $Q(2-\sqrt{6}, 1-\sqrt{6})$ will be $ \begin{aligned} & y-(1-\sqrt{6})=m_2[x-(2-\sqrt{6})]=-1(x-2+\sqrt{6}) \\ & \Rightarrow y-1+\sqrt{6}=-x+2-\sqrt{6} \\ & \Rightarrow x+y=3-2 \sqrt{6} \end{aligned} $

The point $\mathrm{P}(2,1)$ is translated to a point $\mathrm{Q}$ parallel to the line $L \equiv x-y-4=0$ by…

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