Question

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

• A. $(0,1)$
• B. $(0,-1)$
• C. $(4,-1)$
• D. $(-1,4)$

Answer 1

Answer: (C)

Let $A(2,3),B(6,7)$ and $C(8,3)$ be the vertices of parallelogra

And let $D(x,y)$ be the fourth vertex.

Midpoint of $AC$ $=(\frac{2+8}{2},\frac{3+3}{2})=(5,3)$

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

Since the diagonals of parallelogram bisect eachother at $O$

Therefore, midpoint of $AC$ $=$ midpoint of $BD$

$\Rightarrow (\frac{x+6}{2},\frac{y+7}{2})=(5,3)$

$\Rightarrow \frac{x+6}{2}=5,\frac{y+7}{2}=3$

$\Rightarrow x+6=10,y+7=6$

$\Rightarrow x=4,y=-1$

Therefore, the coordinates of the fourth vertex will be $(x,y)=(4,-1)$.