Question

$\text{List I}$ has four entries and $\text{List II}$ has five entries. Each entry of $\text{List I}$ is to be matched with one or more than one entries of $\text{List II}.$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{Centre(s) of the circle(s) having radius}5\text{and}& \text{(P)}& (4,6)\\ & \text{touching the line}3x+4y-11=0\text{at}(1,2)\text{is (are)}& & \\ \text{(B)}& \text{End points of one of the diameters of}& \text{(Q)}& (1,1)\\ & x^2+y^2-6x-8y+20=0\text{are}& & \\ \text{(C)}& \text{The line equidistant from both the lines}& \text{(R)}& (3,-3)\\ & 4x+2y+2=0\text{and}6x+3y-21=0,& & \\ & \text{passes through}& & \\ \text{(D)}& \text{If one of the sides of a square is}3x-4y-12=0& \text{(S)}& (-2,7)\\ & \text{and its centre is}(0,0),\text{then its diagonal}& & \\ & \text{passes through}& & \\ & & \text{(T)}& (-2,-2)\end{array}$

Which of the following is the only CORRECT combination?

• A. $\text{(C)}\to \text{(Q)},\text{(R)},\text{(S)}$
• B. $\text{(C)}\to \text{(Q)},\text{(T)}$
• C. $\text{(D)}\to \text{(P)},\text{(Q)}$
• D. $\text{(D)}\to \text{(R)},\text{(T)}$

Answer 1

The correct option is A $\text{(C)}\to \text{(Q)},\text{(R)},\text{(S)}$

$\left(C\right)$
$4x+2y+2=0\Rightarrow 2x+y+1=0\dots \left(1\right)$
and $6x+3y-21=0\Rightarrow 2x+y-7=0\dots \left(2\right)$
As $\left(1\right)$ and $\left(2\right)$ are parallel,
hence line equidistant from $\left(1\right)$ and $\left(2\right)$ is $2x+y+\left(\frac{1-7}{2}\right)=0$
$\Rightarrow 2x+y-3=0$
By verification $Q,R,S$ can satisfy the above line.

$\left(D\right)$
Slope of $3x-4y-12=0$ is $\frac{3}{4}$
Let $m$ be the slope of a diagonal.
Then $\tan {45}^{\circ }=\left|\frac{m-\frac{3}{4}}{1+\frac{3m}{4}}\right|$
$\Rightarrow m=7$ or $m=-\frac{1}{7}$
$\therefore$ Equations of diagonals are $7x-y=0$ and $x+7y=0$
None of the given points satisfy the above equations.