Question

The point $P(3,6)$ is first reflected on the line $y=x$ and then the image point $Q$ is again reflected on the line $y=-x$ to get the image point ${}^{\prime }{Q}^{\prime }$. Then the circumcentre of the $△PQ{Q}^{\prime }$ is

• A. $(6,3)$
• B. $(6,-3)$
• C. $(3,-6)$
• D. $(0,0)$

Answer 1

The correct option is D $(0,0)$

When the point $P(3,6)$ is reflected on the line $y=x$, the $x$ and $y$ coordinates interchange and we get the point $Q=(6,3)$

Again reflecting $Q^{\prime }$ on the line $y=-x$, we get the the image point $Q^{\prime }=(-3,-6)$

To find out the circumcentre, we need to solve any two bisector equation and then find out the intersection point

So, Midpoint of $P{Q}^{\prime }=(\frac{3-3}{2},\frac{6-6}{2})=(0,0)$

Slope of $P{Q}^{\prime }$ is $m_1=\frac{-6-6}{-3-3}=2$

Slope of bisector is the negative reciprocal of the given slope

So, slope of bisector of $P{Q}^{\prime }$ is $-\frac{1}{2}$

Equation of $P{Q}^{\prime }$ with slope $-\frac{1}{2}$ and point $(0,0)$ is given by

$y-y_1=m(x-x_1)$

$y-0=-\frac{1}{2}(x-0)$

$⟹y=-\frac{1}{2}x$

$⟹2x+y=0$ ........ $\left(i\right)$

Now, Midpoint of $PQ=(\frac{3+6}{2},\frac{6+3}{2})=(\frac{9}{2},\frac{9}{2})$

Slope of $PQ$ is $m_2=\frac{3-6}{6-3}=\frac{-3}{3}=-1$

Slope of bisector of $PQ$ is $1$

$\therefore$ Equation of $PQ$ with slope $1$ and point $(\frac{9}{2},\frac{9}{2})$ is given by

$y-\frac{9}{2}=1(x-\frac{9}{2})$

$⟹y=x$

$⟹x-y=0$ ........... $\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$ we get

$x=0$

Substituting in $\left(i\right)$, we get $y=0$

Therefore, circumcentre of $△PQ{Q}^{\prime }$ is $(0,0)$.