Question

If $P(1,2),Q(a,b),R(5,7)$ and $S(2,3)$ are the vertices of a parallelogram, then sum of the squares of the lengths of its diagonals is

Answer 1

Equation of $PS$ is $y-2=\frac{(2-3)}{(1-2)}(x-1)$ $\Rightarrow$ $x-y+1=0$.

$\therefore$ Equation of $QR$ $($ porallel to $PS)$ is $x-y+\lambda =0$

As this passes through $R($ 5,7 $)$

$\therefore 5-7+\lambda =0\Rightarrow \lambda =2$

Hence equation of $QR$ is $x-y+2=0$ ...(1)

Similarly equation of $SR$ is

$y-3=\frac{3-7}{2-5}(x-2)$ $\Rightarrow 4x-3y+1=0$

$\therefore$ Equation of $PQ$ $($line parallel to $SR)$ is

$4x-3y+\delta =0$

As this passes through $P(1,2)$

$\therefore 4-3\times 2+\delta =0\Rightarrow \delta =2$

$\therefore$ Equation of $PQ$ is $4x-3y+2=0$ ...(2)

Solving (1) and (2) we get coordinates of $Q$ as $(4,6)$

$\therefore$ length of diagonal $PR$ is

$\sqrt{{(5-1)}^2+{(7-2)}^2}=\sqrt{16+25}=\sqrt{41}$

And length of diagonal $SQ$ is

$\sqrt{{(2-4)}^2+{(3-6)}^2}=\sqrt{4+9}=\sqrt{13}$

Hence sum of square of length is of diagonals is $41+13=54$