In a parallelogram, the ends of one diagonal are $\left(3,-4\right)$ and $\left(-6,5\right)$ . If the second diagonal has one end at $\left(-2,1\right)$ then its other end is

Given, in a parallelogram, the ends of one diagonal are $\left(3,-4\right)$ and $\left(-6,5\right)$  and second diagonal are $\left(-2,1\right),\left(2,y\right)$

Let $A=\left(3,-4\right),B=\left(-2,1\right),C=\left(-6,5\right),D=\left(x,y\right)$

In parallelogram diagonals bisect each other

Midpoint of $AC=$ Midpoint of $BD$

$\left(\frac{3-6}{2},\frac{-4+5}{2}\right)=\left(\frac{2-2}{2},\frac{y+1}{2}\right)$

By comparison

$\frac{3-6}{2}=\frac{x-2}{2}$              $\frac{-4+5}{2}=\frac{y+1}{2}$

$x=-1$                        $y=0$

$\therefore D=\left(x,y\right)=\left(-1,0\right)$