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Q.
Column I Column II
A The lines $\left(y-y_1\right)=m\left(x-x_1\right) \pm 4 \sqrt{1+m^2}$ are tangents to the same circle. The radius of the circle is P 4
B A line $L$ passes through a point $P (1,2)$ and has negative gradient. If $L$ makes an angle of $\frac{\pi}{4}$ with the lines $2 x+3 y=10$, the y-intercept of the line $L$ is Q 7
C The line $y=-\frac{3 x}{4}+9$ crosses the $x$-axis at $P$ and the $y$-axis at $Q$. Point $T$ lies on $PQ$ and its coordinates are $( r , s )$. If $\frac{\text { Area of triangle } P O Q}{\text { Area of triangle TOP }}=3$, then the value of $(r+s)$ equals R 8
D Let $ABC$ be a triangle with $AB =3, BC =4$ and $AC =5$. Let $I$ be the centre of the circle inscribed in triangle $ABC$. The product of the distances of incentre from the vertices $A , B$ and $C$ of the triangle $ABC$, is S 10
T 11

Straight Lines

Solution:

(A)image
Now, distance between two parallel lines equals $2 r=8\left(\frac{\sqrt{1+ m ^2}}{\sqrt{1+ m ^2}}\right)=8$
Hence, $r=4$.
(B) Let slope of the line $L$ be $m <0$.
image
Now, $\tan \frac{\pi}{4}=\left|\frac{m-\left(\frac{-2}{3}\right)}{1+m\left(\frac{-2}{3}\right)}\right|$
$\Rightarrow 1=\left|\frac{3 m+2}{3-2 m}\right| \Rightarrow(3 m+2)= \pm(3-2 m)$
Taking positive sign, we get $5 m =1 \Rightarrow m =\frac{1}{5}$
Taking negative sign, we get $ m=-5$
As, $m <0$ (given) so equation of required line is $(y-2)=-5(x-1) \Rightarrow 5 x+y=7$
For $y$-intercept, put $x=0$, we get $y=7$
(C) Line $3 x+4 y=36$
image
$\frac{ x }{12}+\frac{ y }{9}=1 $
$\frac{\Delta_{ OPQ }}{\Delta_{ TOP }}=\frac{54}{6 s }=3 \Rightarrow s =3$
Also $ \frac{ r }{12}+\frac{ s }{9}=1$
$\Rightarrow r + s =11 \text { Ans. ] }$
(D) image
$r =\frac{\Delta}{ s }=\frac{6}{6}=1$
Hence I $(1,1)$
$\therefore AI \cdot BI \cdot CI =(\sqrt{10})(\sqrt{2})(\sqrt{5})=10 $