Question

Let the area of the triangle formed by $(1,2),(-3,0)$ and any point on the line $x-2y+k=0$ is $5$ sq. units. If possible values of $k$ are $k_1,k_2$ then locus of foot of the perpendicular drawn from $(k_1,0)$ to a variable line passing through $(0,k_2)$ is

• A. $x^2+y^2=2x-8y$
• B. $x^2+y^2+2x-8y=0$
• C. $x^2+y^2+8x-2y=0$
• D. $x^2+y^2=8x+2y$

Answer 1

Answer: (B), (C)

$x^2+y^2+8x-2y=0$

Let $A=(1,2),B=(-3,0)$
and locus of variable point be $C=(x,y)$
$△ABC=\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ 1& 2& 1\\ -3& 0& 1\end{array}\right|=5$
$\Rightarrow |(2x-4y+6)|=10\Rightarrow |x-2y+3|=5$
$\Rightarrow x-2y+3=\pm 5$
$\Rightarrow x-2y+8=0\text{or}x-2y-2=0$
$\Rightarrow k_1=-2,k_2=8$ or $k_1=8,k_2=-2$


Case $\left(i\right):k_1=-2,k_2=8$
$\therefore P=(k_1,0)=(-2,0),Q=(0,k_2)=(0,8)$
Slope of $PR\cdot$ slope of $RQ=-1$
$\Rightarrow \frac{y}{x+2}\cdot \frac{y-8}{x}=-1\Rightarrow x(x+2)+y(y-8)=0\Rightarrow x^2+y^2+2x-8y=0$

Case $\left(ii\right):k_1=8,k_2=-2$
Slope of $PR\cdot$ slope of $RQ=-1$
$\Rightarrow \frac{y}{x-8}\cdot \frac{y+2}{x}=-1\Rightarrow x(x-8)+y(y+2)=0\Rightarrow x^2+y^2-8x+2y=0$