Given two points $Q(3,4)$ and $R(1,2)$. What is the point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ on the line…
Given two points $Q(3,4)$ and $R(1,2)$. What is the point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ on the line $2 \mathrm{x}-\mathrm{y}-1=0$ for which $\mathrm{PQ}+\mathrm{PR}=\mathrm{QR}$ holds
$(-3,-7)$
$(-2,-5)$
$(2,3)$
$(4,7)$
Solution
We are given that
$\mathrm{PQ}+\mathrm{PB}=\mathrm{QR}$
$\begin{aligned} & \Rightarrow \sqrt{(\mathrm{x}-3)^2+(\mathrm{y}-4)^2}+\sqrt{(\mathrm{x}-1)^2+(\mathrm{y}-2)^2} \\ & \quad=\sqrt{(3-1)^2+(4-7)^2} \\ & \Rightarrow \sqrt{(\mathrm{x}-3)^2+(\mathrm{y}-4)^2}+\sqrt{(\mathrm{x}-1)^2+(\mathrm{y}-2)^2}=\sqrt{4+4} \\ & \Rightarrow \sqrt{(\mathrm{x}-3)^2+(\mathrm{y}-4)^2}+\sqrt{(\mathrm{x}-1)^2+(\mathrm{y}-2)^2}=\sqrt{8} \\ & \Rightarrow \sqrt{(2-3)^2+(3-4)^2}+\sqrt{(2-1)^2+(3-2)^2} \\ & \Rightarrow \sqrt{1+1}+\sqrt{1+1}=2 \sqrt{2}=\sqrt{8}\end{aligned}$