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

Mid-point of $\mathrm{P}(2,3)$ and $\mathrm{Q}(4,5)=(3,4)$ Slope of PQ = 1 Slope of the line $\mathrm{L}=-1$ Mid-point $(3,4)$ lies on the line $L$. Equation of line $\mathrm{L}$, y-4=-1(x-3) \Rightarrow x+y-7=0 Let image$of point\mathrm{R}(0,0)$ be $\mathrm{S} \left(x_1, y_1\right) Mid-point$ of $\mathrm{RS}$=\left(\frac{x_1}{2}, \frac{y_1}{2}\right)$Mid-point$\left(\frac{x_1}{2}, \frac{y_1}{2}\right)$lies on the line (i)$x_1+y_1=14Slope of $RS =$\frac{y_1}{x_1} Since \mathrm{RS} \perp$line$\mathrm{L}$\begin{aligned} \frac{y_1}{x_1} \times(-1)=-1 \\ &x_1=y_1 \end{aligned}$$$ From (ii) and (iii), x_1=y_1=7 Hence the image of R=(7,7)