Is it true that the points $A (2, − 2), B (14, 10), C (11, 13)$ and $D (− 1, 1)$ and are the vertices of a rectangle?

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True.

False

answer is A.

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Detailed Solution

We have been given points,

A (2, − 2), B (14, 10), C (11, 13)

and

D (− 1, 1)

.
Using the distance formula,

D = x 2 − x 1 2 + y 2 − y 1 2

As we know, in a rectangle, the lengths of the diagonals and the opposing sides are equal.
Calculating the lengths of the sides by using the distance formula,
For AB substituting

x_{2} = 14, x_{1} = 2, y_{2} = 10, and y_{1} = - 2

, in the formula,

AB = \sqrt{{(14 - 2)}^{2} + {(10 + 2)}^{2}} = 12 \sqrt{2}

For BC substituting

x_{2} = 11, x_{1} = 14, y_{2} = 13, and y_{1} = 10

, in the formula,

BC = \sqrt{{(11 - 14)}^{2} + {(13 - 10)}^{2}} = 3 \sqrt{2}

For DC substituting

x_{2} = 11, x_{1} = - 1, y_{2} = 13, and y_{1} = 1

, in the formula,

DC = \sqrt{{(11 + 1)}^{2} + {(13 - 1)}^{2}} = 12 \sqrt{2}

For DA substituting

x_{2} = 2, x_{1} = - 1, y_{2} = - 2, and y_{1} = 1

, in the formula,

DA = \sqrt{{(2 + 1)}^{2} + {(- 2 - 1)}^{2}} = 3 \sqrt{2}

For AC substituting

x_{2} = 11, x_{1} = 2, y_{2} = 13, and y_{1} = - 2

, in the formula,

AC = \sqrt{{(11 - 2)}^{2} + {(13 + 2)}^{2}} = \sqrt{306}

For BD substituting

x_{2} = 14, x_{1} = - 1, y_{2} = 10, and y_{1} = 1

, in the formula,

BD = \sqrt{{(14 + 1)}^{2} + {(10 - 1)}^{2}} = \sqrt{306}

In the rectangle, the diagonals' lengths and the lengths of the opposite sides are both equal. It is therefore a rectangle.
Therefore, our statement is true.
Hence, option 1 is correct.

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