Question

Let P point on the circle $x^2+y^2=9$, Q a point on the line $7x+y+3=0$, and the perpendicular bisector of PQ be the line $x-y+1=0$. Then the coordinate of P are

• A. (0, -3)
• B. (0, 3)
• C. $(\frac{72}{25},-\frac{21}{25})$
• D. $(-\frac{72}{25},\frac{21}{25})$

Answer 1

The correct option is D $(-\frac{72}{25},\frac{21}{25})$

Any point on the lines $7x+y+3=0$ is $Q(t,-3-7t),t\in R.$
Now P $(h,k)$ is image of point Q in the line $x-y+1=0$
Then, $\frac{h-t}{1}=\frac{k-(-3-7t)}{-1}$
$=-\frac{2(t-(-3-7t)+1)}{1+1}$
$=-8t-4$
$\Rightarrow (h,k)\equiv (-7t-4,t+1)$
This point lies on the circle $x^2+y^2=9$
$\Rightarrow (-7t-4{)}^2+(t+1{)}^2=9$
$\Rightarrow 50r^2+58t+8=0$
$\Rightarrow 25r^2+19t+4=0$
$\Rightarrow (25t+4)(t+1)=0$
$\Rightarrow t=-4/25,t=1$
$\Rightarrow (h,k)=(-\frac{72}{25},\frac{21}{25})$ or $(3,0)$