Question

Prove that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices of a rectangle.

Answer 1

Hint: Firstly, we will find the lengths of all the sides of the rectangle and show that opposite sides are equal and then find the slopes and show that opposite sides are parallel. Thus, we have a parallelogram. Now, we just need to show two adjacent sides are perpendicular to each other.

Complete step-by-step answer:
To prove that a quadrilateral is a rectangle we need to show that opposite sides are equal and parallel and it has at least one right angle.
Let A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) be the given vertices. Now we need to show that ABCD is a rectangle.
First, let us find the side lengths of AB, BC, CD, AD.
We know that distance between two points (a, b) and (c, d) is given by the formula:
$d=\sqrt{{(c-a)}^2+{(d-b)}^2}$
We have A(-4, -1) and B(-2, -4).
So, $AB=\sqrt{{(-2+4)}^2+{(-4+1)}^2}$
$AB=\sqrt{{(2)}^2+{(-3)}^2}=\sqrt{4+9}=\sqrt{13}$ ……(1)
We have B(-2, -4) and C(4, 0).
So, $BC=\sqrt{{(4+2)}^2+{(0+4)}^2}$
$BC=\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}$ ……….(2)
We have C(4, 0) and D(2, 3).
So, $CD=\sqrt{{(2-4)}^2+{(3-0)}^2}$
$CD=\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13}$ ……….(3)
We have A(-4, -1) and D(2, 3).
So, $AD=\sqrt{{(2+4)}^2+{(3+1)}^2}$
$AD=\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}$ ……….(4)
Now comparing (1) and (3), we have:- AB = CD.
And comparing (2) and (4), we have:- BC = AD.
Hence, opposite sides are equal in length.
Now, we will find the slopes of the all the sides:-
Slope of line formed by two points (a, b) and (c, d) is given by ${\displaystyle \frac{d-b}{c-a}}$.
We have A(-4, -1) and B(-2, -4).
So, Slope of $AB={\displaystyle \frac{-4+1}{-2+4}}=-{\displaystyle \frac{3}{2}}$ ……..(5)
We have B(-2, -4) and C(4, 0).
So, Slope of $BC={\displaystyle \frac{0+4}{4+2}}={\displaystyle \frac{4}{6}}={\displaystyle \frac{2}{3}}$ ………(6)
We have C(4, 0) and D(2, 3).
So, Slope of $CD={\displaystyle \frac{3-0}{2-4}}={\displaystyle \frac{3}{-2}}=-{\displaystyle \frac{3}{2}}$ ………(7)
We have D(2, 3) and A(-4, -1).
So, Slope of $DA={\displaystyle \frac{-1-3}{-4-2}}={\displaystyle \frac{-4}{-6}}={\displaystyle \frac{2}{3}}$ ………(8)
Now comparing (5) and (7), we have:- Slope of AB = Slope of CD.
And comparing (6) and (8), we have:- Slope of BC = Slope of AD.
Hence, opposite sides are parallel.
Hence, now, ABCD is a parallelogram.
(Because If a quadrilateral has equal opposite sides and parallel, then it is a parallelogram)
Now we can see from (5) and (6) that Slope of AB is a negative reciprocal of Slope of BC.
This happens when lines are perpendicular.
Similarly with (7) and (8).
Hence, we got a rectangle by the requirements mentioned in the start of the answer.

Note: The students might make the mistake of forgetting any one of the conditions of proving a quadrilateral is a rectangle like maybe he/she forgets to prove that opposite sides are parallel or equal.
Or he might stop at proving ABCD a parallelogram, but we need one right angle for sure.