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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 $\dfrac{{d - b}}{{c - a}}$.

We have A(-4, -1) and B(-2, -4).

So, Slope of $AB = \dfrac{{ - 4 + 1}}{{ - 2 + 4}} = - \dfrac{3}{2}$ ……..(5)

We have B(-2, -4) and C(4, 0).

So, Slope of $BC = \dfrac{{0 + 4}}{{4 + 2}} = \dfrac{4}{6} = \dfrac{2}{3}$ ………(6)

We have C(4, 0) and D(2, 3).

So, Slope of $CD = \dfrac{{3 - 0}}{{2 - 4}} = \dfrac{3}{{ - 2}} = - \dfrac{3}{2}$ ………(7)

We have D(2, 3) and A(-4, -1).

So, Slope of $DA = \dfrac{{ - 1 - 3}}{{ - 4 - 2}} = \dfrac{{ - 4}}{{ - 6}} = \dfrac{2}{3}$ ………(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.

Or he might stop at proving ABCD a parallelogram, but we need one right angle for sure.