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. (-1, 0)
• D. (1, 0

Answer 1

The correct option is B (0, -1)

Let the fourth vertex D $=(x,y)$
We know that the diagonals of a parallelogram bisect each other. So,the
midpoint of AC is same as the mid point of BD.
Mid point of two points ${(x}_1,y_1)$ and ${(x}_2,y_2)$ is calculated by the formula $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
So, midpoint of $AC=$ Mid point of $BD$
$=>(\frac{-2+8}{2},\frac{3+3}{2})=(\frac{6+x}{2},\frac{7+y}{2})$
$=>(\frac{6}{2},\frac{6}{2})=(\frac{6+x}{2},\frac{7+y}{2})$
$=>6+x=6;7+y=6$
$=>x=0;y=-1$
Hence, $D=(0,-1)$