Question

Show that the following points taken in order form the vertices of a parallelogram

(3, -5), (-5, -4), (7, 10) and (15, 9)

Answer 1

Let $A,B,C$ and $D$ represent the points $(3,–5),(–5,-4),(7,10)$ and $(15,9)$

Distance of $AB$

$AB=\sqrt{(–5-3{)}^2+(-4+5{)}^2}$

$AB=\sqrt{(-8{)}^2+1^2}$

$AB=\sqrt{64+1}$

$AB=\sqrt{65}$

Similarly

Distance of $BC$

$BC=\sqrt{(7+5{)}^2+(10+4{)}^2}$

$BC=\sqrt{{12}^2+{14}^2}$

$BC=\sqrt{144+196}$

$BC=\sqrt{340}$

Similarly

Distance of $CD$

$CD=\sqrt{(15-7{)}^2+(9-10{)}^2}$

$CD=\sqrt{(8{)}^2+(-1{)}^2}$

$CD=\sqrt{64+1}$

$CD=\sqrt{65}$

Similarly

Distance of $DA$

$DA=\sqrt{(3-15{)}^2+(-5-9{)}^2}$

$DA=\sqrt{(-12{)}^2+(-14{)}^2}$

$DA=\sqrt{144+196}$

$DA=\sqrt{340}$

So, $AB=CD=\sqrt{65}$ and $BC=DA=\sqrt{340}$

i.e., The opposite sides are equal.

Hence $ABCD$ is a parallelogram.