Suppose P and Q are the mid points of the sides A B and B C of a triangle where A 1 , 3 ,   B 3 , 7 and…

Given,

$P$ and $Q$ are the mid points of the sides $AB$ and $BC$ of a triangle where $A\left(1,3\right), B\left(3,7\right)$ and $C\left(7,15\right)$ are vertices,

So, coordinates of $P$ are $\left(\frac{3+1}{2},\frac{7+3}{2}\right)=\left(2,5\right)$ and of $Q$ are $\left(\frac{3+7}{2},\frac{7+15}{2}\right)=\left(5,11\right)$.

Let the coordinates of $R$ be $\left(x,y\right)$,

Then $A{C}^2+Q{R}^2=P{R}^2$

 $\Rightarrow {\left(PR\right)}^2-{\left(QR\right)}^2={\left(AC\right)}^2$

$\Rightarrow {\left(\mathrm{x}-2\right)}^2+{\left(\mathrm{y}-5\right)}^2-\left[{\left(\mathrm{x}-5\right)}^2+{\left(\mathrm{y}-11\right)}^2\right]=\left[{\left(7-1\right)}^2+{\left(15-3\right)}^2\right]$

$\Rightarrow 6x+12y+4-121=36+144$

$\Rightarrow 6x+12y=297$ which is the locus of $R$