Question

prove that the vertices of a(-2, 3), b(4, 3), c(4, -1) and d(-2, 1) form a rectangle​

Answer 1

Answer:

In a rectangle opposite sides are equal

Since,

The opposite sides are equal in this figure

Therefore,

ABCD is a rectangle

Answer 2

Answer:

Solution:

$\mathsf{AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

$\mathsf{AB=\sqrt{(-2-4)^2+(3-3)^2}}$

$\mathsf{AB=\sqrt{(-6)^2+(0)^2}}$

$\mathsf{AB=\sqrt{36}}$

$\implies\boxed{\mathsf{AB=6}}$

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

$\mathsf{BC=\sqrt{(0)^2+(4)^2}}$

$\mathsf{BC=\sqrt{16}}$

$\implies\boxed{\mathsf{BC=4}}$

$\mathsf{CD=\sqrt{(4+2)^2+(-1+1)^2}}$

$\mathsf{CD=\sqrt{6^2+(0)^2}}$

$\mathsf{CD=\sqrt{36}}$

$\implies\boxed{\mathsf{CD=6}}$

$\mathsf{AD=\sqrt{(-2+2)^2+(3+1)^2}}$

$\mathsf{AD=\sqrt{(0)^2+(4)^2}}$

$\mathsf{AD=\sqrt{16}}$

$\implies\boxed{\mathsf{AD=4}}$

$\mathsf{AB=CD\;\;\;and\;\;BC=AD}$

$\implies\textsf{Opposite sides are equal}$

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

$\mathsf{AC=\sqrt{(-6)^2+4^2}}$

$\mathsf{AC=\sqrt{36+16}}$

$\mathsf{AC=\sqrt{5}}$

$\mathsf{Now,}$

$\mathsf{AB^2+BC^2}$

$\mathsf{=36+16}$

$\mathsf{=52}$

$\mathsf{=AC^2}$

$\implies\mathsf{\angle{B}=90^\circ}$

$\textsf{Hence ABCD is a rectangle}$

hope it helps ✔︎✔️