Question

The fourth vertex of D of a parallelogram ABCD whose three vertices are $A(-2,5)B(6,7)$ and $C(8,3)$ is

• A. $(-1,0)$
• B. $(1,0)$
• C. $(0,-1)$
• D. $(0,1)$

Answer 1

The correct option is D $(0,1)$

Let D(x,y) is the fourth vertex of the parallelogram $ABCD.$

Mid point of AC$=(\frac{-2+8}{2},\frac{5+3}{2})=(\frac{6}{2},4)$

Mid point of BD$=(\frac{6+x}{2},\frac{7+y}{2})$

Since, the diagonal of the parallelogram bisect each other at O

$\therefore$Mid point of BD=Mid point of AC

$(\frac{6+x}{2},\frac{7+y}{2})=(\frac{6}{2},4)$

$\Rightarrow \frac{6+x}{2}=\frac{6}{2},\frac{7+y}{2}=4$

$\Rightarrow 12+2x=12,7+y=8$

$\Rightarrow 2x=0,y=8-7$

$\Rightarrow x=0,y=1$