Question

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

Answer 1

Let the points are $A(-7,-3),B(5,10),C(15,8),D(3,-5),$

$\begin{array}{cc}AB& =\sqrt{{(5-(-7\left)\right)}^2+{(10-(-3\left)\right)}^2}\\ & =\sqrt{{\left(12\right)}^2+{\left(13\right)}^2}\\ & =\sqrt{144+169}\\ & =\sqrt{313}\end{array}$

$\begin{array}{cc}BC& =\sqrt{{(15-5)}^2+{(8-10)}^2}\\ & =\sqrt{{\left(10\right)}^2+2^2}\\ & =\sqrt{100+4}\\ & =\sqrt{104}\end{array}$

$\begin{array}{cc}CD& =\sqrt{{(3-15)}^2+{(-5-8)}^2}\\ & =\sqrt{{\left(12\right)}^2+{\left(13\right)}^2}\\ & =\sqrt{144+169}\\ & =\sqrt{313}\end{array}$

$\begin{array}{cc}DA& =\sqrt{{(-7-3)}^2+{(-3-(-5\left)\right)}^2}\\ & =\sqrt{{\left(10\right)}^2+2^2}\\ & =\sqrt{100+4}\\ & =\sqrt{104}\end{array}$

Now, we will check the relation between diagonals,

$\begin{array}{cc}AC& =\sqrt{{(15-(-7\left)\right)}^2+{(8-(-3\left)\right)}^2}\\ & =\sqrt{{\left(22\right)}^2+{\left(11\right)}^2}\\ & =\sqrt{484+121}\\ & =\sqrt{605}\end{array}$

$\begin{array}{cc}BD& =\sqrt{{(3-5)}^2+{(-5-10)}^2}\\ & =\sqrt{2^2+{\left(15\right)}^2}\\ & =\sqrt{4+225}\\ & =\sqrt{229}\end{array}$

Since , $AB=CD,BC=DA,AC\ne BD$. Adjacent sides are not equal as well hence it is a parallelogram.