Question

Prove that , the vertices of A(-2,3) ,B(4,3), C(4,-1) and D(-2,-1) form a rectangle. ​

Answer 1

Given :-

A quadrilateral ABCD having coordinates

• A(-2,3)

• B(4,3)

• C(4,-1)

• D(-2,-1)

To Show :-

• ABCD is a rectangle.

Concept Used :-

In order to show that ABCD is a rectangle, its enough to show that

• AB = CD

• AD = BC

• AC = BD

Formula Used :-

Distance Formula :-

Let us consider a line segment joining the points A and B, then distance between A and B is

${\underline{\boxed{\rm{\quad Distance \: AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$

Solution :-

Given that,

A quadrilateral ABCD having coordinates

• A(-2,3)

• B(4,3)

• C(4,-1)

• D(-2,-1)

So,

$\rm :\longmapsto\:AB = \sqrt{ {(4 + 2)}^{2} + {(3 - 3)}^{2} }$

$\rm :\longmapsto\:AB = \sqrt{ {(6)}^{2} + {(0)}^{2} }$

$\rm :\longmapsto\:AB = \sqrt{ {(6)}^{2}}$

$\bf\implies \:AB = 6 \: units$

$\rm :\longmapsto\:BC = \sqrt{ {(4 - 4)}^{2} + {( - 1 - 3)}^{2} }$

$\rm :\longmapsto\:BC = \sqrt{ {(0)}^{2} + {( -4)}^{2} }$

$\rm :\longmapsto\:BC = \sqrt{ {(4)}^{2} }$

$\bf\implies \:BC = 4 \: units$

$\rm :\longmapsto\:CD = \sqrt{ {( - 4 - 2)}^{2} + {( - 1 + 1)}^{2} }$

$\rm :\longmapsto\:CD = \sqrt{ {( - 6)}^{2} + {(0)}^{2} }$

$\rm :\longmapsto\:CD = \sqrt{ {(6)}^{2} }$

$\bf\implies \:CD = 6 \: units$

$\rm :\longmapsto\:DA = \sqrt{ {( - 2 + 2)}^{2} + {(3 + 1)}^{2} }$

$\rm :\longmapsto\:DA = \sqrt{ {(0)}^{2} + {(4)}^{2} }$

$\rm :\longmapsto\:DA = \sqrt{ {(4)}^{2} }$

$\bf\implies \:DA = 4 \: units$

$\rm :\longmapsto\:AC = \sqrt{ {(4 + 2)}^{2} + {( - 1 - 3)}^{2} }$

$\rm :\longmapsto\:AC = \sqrt{ {(6)}^{2} + {( - 4)}^{2} }$

$\rm :\longmapsto\:AC = \sqrt{ 36 + 16}$

$\rm :\longmapsto\:AC = \sqrt{52}$

$\bf\implies \:\:AC = 2\sqrt{13} \: units$

$\rm :\longmapsto\:BD = \sqrt{ {(4 + 2)}^{2} + {( - 1 - 3)}^{2} }$

$\rm :\longmapsto\:BD = \sqrt{ {(6)}^{2} + {( - 4)}^{2} }$

$\rm :\longmapsto\:BD = \sqrt{36 + 16}$

$\rm :\longmapsto\:BD = \sqrt{52}$

$\bf\implies \:BD = 2 \sqrt{13} \: units$

Hence,

$\rm :\longmapsto\:AB = CD \\ \rm :\longmapsto\:BC = DA \\ \rm :\longmapsto\:AC = BD$

Hence,

• ABCD is a rectangle.

Additional Information :-

Midpoint Formula :-

Let us consider a line segment joining the points A and B and let C (x, y) be the midpoint of AB, then coordinates of C is

$\boxed{ \quad\sf \:( x, y) = \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg) \quad}</p><p>$

$\bf \ Area =\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Section Formula :-

Let us consider a line segment joining the points A and B and let C (x, y) be the point on AB, which divides AB in the ration m : n, then coordinates of C is

$\boxed{ \quad\sf \:( x, y) = \bigg(\dfrac{nx_1+mx_2}{m + n} , \dfrac{ny_1+my_2}{m + n} \bigg) \quad}</p><p>$