The reflection of the point (1, 1) along the | vedclass

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(C) Let the reflection of the point $P(1, 1)$ be $Q(h, k)$.The line of reflection is $x + y = 0$.The mid-point of $PQ$ is $\left(\frac{h+1}{2}, \frac{

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$(-1, -1)$

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(C) Let the reflection of the point $P(1, 1)$ be $Q(h, k)$.The line of reflection is $x + y = 0$.The mid-point of $PQ$ is $\left(\frac{h+1}{2}, \frac{k+1}{2}ight)$,which must lie on the line $x + y = 0$.$\frac{h+1}{2} + \frac{k+1}{2} = 0 \Rightarrow h + k + 2 = 0 \quad \dots(i)$The line $PQ$ is perpendicular to the line $x + y = 0$ (which has slope $-1$).The slope of $PQ$ is $\frac{k-1}{h-1}$.Since $PQ$ is perpendicular to $x + y = 0$,the product of their slopes is $-1$.$\left(\frac{k-1}{h-1}ight) 	imes (-1) = -1$ $\Rightarrow \frac{k-1}{h-1} = 1$ $\Rightarrow k - 1 = h - 1$ $\Rightarrow k = h$.Substituting $k = h$ into equation $(i)$:$h + h + 2 = 0$ $\Rightarrow 2h = -2$ $\Rightarrow h = -1$.Since $k = h$,we have $k = -1$.Thus,the reflection of the point $(1, 1)$ along the line $y = -x$ is $(-1, -1)$.

The reflection of the point $(1, 1)$ along the line $y = -x$ is

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