If the image of point P(2, 3) in a line L is | vedclass

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(D) Mid-point of $P(2, 3)$ and $Q(4, 5)$ is $(\frac{2+4}{2}, \frac{3+5}{2}) = (3, 4)$.Slope of $PQ = \frac{5-3}{4-2} = \frac{2}{2} = 1$.Since the line

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$(7, 7)$

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(D) Mid-point of $P(2, 3)$ and $Q(4, 5)$ is $(\frac{2+4}{2}, \frac{3+5}{2}) = (3, 4)$.Slope of $PQ = \frac{5-3}{4-2} = \frac{2}{2} = 1$.Since the line $L$ is the perpendicular bisector of $PQ$,the slope of line $L$ is $m = -1$.The equation of line $L$ passing through $(3, 4)$ with slope $-1$ is $y - 4 = -1(x - 3) \Rightarrow x + y - 7 = 0$.Let the image of point $R(0, 0)$ be $S(x_1, y_1)$.The mid-point of $RS$ is $(\frac{x_1}{2}, \frac{y_1}{2})$,which lies on line $L$,so $\frac{x_1}{2} + \frac{y_1}{2} - 7 = 0 \Rightarrow x_1 + y_1 = 14$.The slope of $RS$ is $\frac{y_1}{x_1}$. Since $RS \perp L$,the slope of $RS$ must be $1$ (negative reciprocal of $-1$).Thus,$\frac{y_1}{x_1} = 1 \Rightarrow x_1 = y_1$.Substituting $x_1 = y_1$ into $x_1 + y_1 = 14$,we get $2x_1 = 14 \Rightarrow x_1 = 7$ and $y_1 = 7$.Therefore,the image of $R$ is $(7, 7)$.

If the image of point $P(2, 3)$ in a line $L$ is $Q(4, 5)$,then the image of point $R(0, 0)$ in the same line is

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