Question

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 $2x+3y=10$, the y-intercept of the line $L$ is	Q	7
C	The line $y=-\frac{3x}{4}+9$ crosses the $x$-axis at $P$ and the $y$-axis at $Q$. Point $T$ lies on $PQ$ and its coordinates are $(r,s)$. If $\frac{\text{ Area of triangle }POQ}{\text{ Area of triangle TOP }}=3$, 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

• A. (A) P ; (B) Q ; (C) T; (D) S
• B. (A) Q; (B) Q ; (C) T; (D) S
• C. (A) R; (B) Q ; (C) T; (D) S
• D. (A) S; (B) Q ; (C) T; (D) S

Answer 1

Answer: (A)

(A)
Now, distance between two parallel lines equals $2r=8\left(\frac{\sqrt{1+m^2}}{\sqrt{1+m^2}}\right)=8$
Hence, $r=4$.
(B) Let slope of the line $L$ be $m<0$.

Now, $\tan \frac{\pi }{4}=\left|\frac{m-\left(\frac{-2}{3}\right)}{1+m\left(\frac{-2}{3}\right)}\right|$
$\Rightarrow 1=\left|\frac{3m+2}{3-2m}\right|\Rightarrow (3m+2)=\pm (3-2m)$
Taking positive sign, we get $5m=1\Rightarrow m=\frac{1}{5}$
Taking negative sign, we get $m=-5$
As, $m<0$ (given) so equation of required line is $(y-2)=-5(x-1)\Rightarrow 5x+y=7$
For $y$-intercept, put $x=0$, we get $y=7$
(C) Line $3x+4y=36$

$\frac{x}{12}+\frac{y}{9}=1$
$\frac{{\Delta }_{OPQ}}{{\Delta }_{TOP}}=\frac{54}{6s}=3\Rightarrow s=3$
Also $\frac{r}{12}+\frac{s}{9}=1$
$\Rightarrow r+s=11\text{ Ans. ] }$
(D)
$r=\frac{\Delta }{s}=\frac{6}{6}=1$
Hence I $(1,1)$
$\therefore AI\cdot BI\cdot CI=(\sqrt{10})(\sqrt{2})(\sqrt{5})=10$