Let Q be the image of a point P(1, 2) with re | vedclass

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(A) The image $Q(x_1, y_1)$ of point $P(1, 2)$ with respect to $x + y + 1 = 0$ is given by $\frac{x_1 - 1}{1} = \frac{y_1 - 2}{1} = -2 \frac{1(1) + 1(

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$\sqrt{10}$

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(A) The image $Q(x_1, y_1)$ of point $P(1, 2)$ with respect to $x + y + 1 = 0$ is given by $\frac{x_1 - 1}{1} = \frac{y_1 - 2}{1} = -2 \frac{1(1) + 1(2) + 1}{1^2 + 1^2} = -2 \frac{4}{2} = -4$.
Thus,$x_1 - 1 = -4 \implies x_1 = -3$ and $y_1 - 2 = -4 \implies y_1 = -2$. So,$Q = (-3, -2)$.
$M$ is the midpoint of $PQ$,so $M = (\frac{1-3}{2}, \frac{2-2}{2}) = (-1, 0)$.
The image $R(x_2, y_2)$ of point $Q(-3, -2)$ with respect to $x - y - 1 = 0$ is given by $\frac{x_2 + 3}{1} = \frac{y_2 + 2}{-1} = -2 \frac{1(-3) - 1(-2) - 1}{1^2 + (-1)^2} = -2 \frac{-3 + 2 - 1}{2} = -2 \frac{-2}{2} = 2$.
Thus,$x_2 + 3 = 2 \implies x_2 = -1$ and $y_2 + 2 = -2 \implies y_2 = -4$. So,$R = (-1, -4)$.
$N$ is the midpoint of $QR$,so $N = (\frac{-3-1}{2}, \frac{-2-4}{2}) = (-2, -3)$.
The distance $MN = \sqrt{(-2 - (-1))^2 + (-3 - 0)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$.

Let $Q$ be the image of a point $P(1, 2)$ with respect to the line $x + y + 1 = 0$ and $R$ be the image of $Q$ with respect to the line $x - y - 1 = 0$. If $M$ and $N$ are the midpoints of $PQ$ and $QR$ respectively,then $MN =$

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