Question

Assertion :Reflections of the point $(-3,2)$ in the line $x+y=0$ is $(-2,3).$ Reason: The reflection of a point $P(\alpha ,\beta )$ in the line $ax+by+c=0$ is the point $Q({\alpha }^{\prime },{\beta }^{\prime })$ if $(\frac{\alpha +{\alpha }^{\prime }}{2},\frac{\beta +{\beta }^{\prime }}{2})$ lies on the line.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

The reflection of a points $P(\alpha ,\beta )$ in the line $ax+by+c=0$ is the point $Q\left(\alpha {,}^{\prime }\beta {,}^{\prime }\right)$ such that the mid-pointy of $PQ$, i.e. $(\frac{\alpha +{\alpha }^{\prime }}{2},\frac{\beta +{\beta }^{\prime }}{2})$ lies on the line.

$PQ$ is perpendicular to the line $ax+by+c=0$.

Also, the coordinates of $Q$ are given by $\frac{\alpha -{\alpha }^{\prime }}{a}=\frac{{\beta }^{\prime }-\beta }{b}=-2\left(\frac{a\alpha +b\beta +c}{a^2+b^2}\right)$

So, Statement $II$ is not true.

The reflection $(-3,2)$ in the line $x+y=0$ has the coordinates of $(x,y)$ given by $\frac{x+3}{1}=\frac{y-2}{1}=-\frac{2(-3+2)}{1^2+1^2}$

$\Rightarrow x+3=y-2=1\Rightarrow x=-2$ and $y=3$

So, Statement $I$ is true.