Question

Prove that the points P (0, -4), Q (6, 2),
R (3, 5) and S (-3, -1) are the vertices of a
rectangle PQRS.

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Answer 1

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Answer 2

The points P (0, -4), Q (6, 2),  R (3, 5) and S (-3, -1) are the vertices of a  rectangle PQRS, proved.

Step-by-step explanation:

Given,

The four points P (0, - 4), Q (6, 2),  R (3, 5) and S (- 3, - 1) are the vertices of a rectangle PQRS.

Using distance formula,

$\sqrt{(x_{2}-x_{1} )^{2}+(y_{2}-y_{1} )^{2}}$

∴ PQ = $\sqrt{(6-0)^{2}+(2+4)^{2}}$

$=\sqrt{(6)^{2}+(6)^{2}} =\sqrt{36+36}$ = $\sqrt{72}$

QR = $\sqrt{(3-6)^{2}+(5-2)^{2}}$

$=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}$

RS = $\sqrt{(-3-3)^{2}+(-1-5)^{2}}$

$=\sqrt{(-6)^{2}+(-6)^{2}}=\sqrt{36+36} =\sqrt{72}$

SP = $\sqrt{(-3-0)^{2}+(-1+4)^{2}}$

$=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9} =\sqrt{18}$

∴ PQ = RS = $\sqrt{72}$ and

QR = SP = $\sqrt{18}$

Diagonals,

PR =$\sqrt{(3-0)^{2}+(5+4)^{2}}$

$=\sqrt{(3)^{2}+(9)^{2}}=\sqrt{9+81} =\sqrt{90}$

QS = $\sqrt{(-3-6)^{2}+(-1-2)^{2}}$

$=\sqrt{(-9)^{2}+(-3)^{2}}=\sqrt{81+9} =\sqrt{90}$

Diagonal PR = Diagonal QS = $\sqrt{90}$, proved.

Thus, the points P (0, -4), Q (6, 2),  R (3, 5) and S (-3, -1) are the vertices of a  rectangle PQRS, proved.