If Q and R are the images of the point P(2,3) | vedclass

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(C) Let $Q = (h, k)$. The image of $P(2, 3)$ with respect to $x-y+2=0$ is given by $\frac{h-2}{1} = \frac{k-3}{-1} = -2 \frac{2-3+2}{1^2+(-1)^2} = -2

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the same side of the line $x+y+2=0$

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(C) Let $Q = (h, k)$. The image of $P(2, 3)$ with respect to $x-y+2=0$ is given by $\frac{h-2}{1} = \frac{k-3}{-1} = -2 \frac{2-3+2}{1^2+(-1)^2} = -2 \frac{1}{2} = -1$. Thus,$h-2 = -1 \Rightarrow h=1$ and $k-3 = 1 \Rightarrow k=4$. So,$Q = (1, 4)$. Let $R = (x_1, y_1)$. The image of $P(2, 3)$ with respect to $2x+y-2=0$ is given by $\frac{x_1-2}{2} = \frac{y_1-3}{1} = -2 \frac{2(2)+3-2}{2^2+1^2} = -2 \frac{5}{5} = -2$. Thus,$x_1-2 = -4 \Rightarrow x_1=-2$ and $y_1-3 = -2 \Rightarrow y_1=1$. So,$R = (-2, 1)$. Now,check the position of $Q(1, 4)$ and $R(-2, 1)$ with respect to the line $x+y+2=0$: For $Q(1, 4)$,$1+4+2 = 7 > 0$. For $R(-2, 1)$,$-2+1+2 = 1 > 0$. Since both values are positive,$Q$ and $R$ lie on the same side of the line $x+y+2=0$.

If $Q$ and $R$ are the images of the point $P(2,3)$ with respect to the lines $x-y+2=0$ and $2x+y-2=0$ respectively,then $Q$ and $R$ lie on

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