Question

Prove that the points $(-7,-3),(5,10),(15,8)$ and $(3,-5)$ taken in order are the vertices of a parallelogram.

Answer 1

Let the given points be $A(-7,-3),B(5,10),C(15,8),D(3,-5)$, taken in order.

We know that, the diagonals of a parallelogram bisect each other.

Hence, $ABCD$ will be a parallelogram if,

Midpoint of $AC=$ Midpoint of $BD$

We also know that, the coordinates of the midpoint of the line segment joining $(x_1,y_1)and(x_2,y_2)$ are:

$P(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Hence, coordinates of midpoint of $AC=(\frac{-7+15}{2},\frac{-3+8}{2})$

$=(4,\frac{5}{2})$

And, coordinates of midpoint of $BD=(\frac{5+3}{2},\frac{10-5}{2})$

$=(4,\frac{5}{2})$

Since, the coordinates of the midpoint of $AC$ and $BD$ are same, the $ABCD$ must form a parallelogram.

Hence, points $A,B,C,D$ are the vertices of a parallelogram.