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  • The points A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0) in that order form a rectangle.

The points A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0) in that order form a rectangle.

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Answer.    [True]
Solution.   The given points are A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0)
Length \, of \, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}
AB= \sqrt{\left ( 4+1 \right )^{2}+\left ( 3+2 \right )^{2}}
AB= \sqrt{50}= 5\sqrt{2}
Length \, of \, BC= \sqrt{\left ( 2-4 \right )^{2}+\left ( 5-3 \right )^{2}}
BC=\sqrt{4+4}= \sqrt{8}= 2\sqrt{2}
Length \, of \, CD= \sqrt{\left ( -3-2 \right )^{2}+\left ( 0-5 \right )^{2}}
CD= \sqrt{25+25}= \sqrt{50}= 5\sqrt{2}
Length \, of \, DA= \sqrt{\left ( -1+3 \right )^{2}+\left ( -2-0 \right )^{2}}
DA= \sqrt{4+4}= \sqrt{8}= 2\sqrt{2}
AB = CD , BC = DA
Length \, of \, AC= \sqrt{\left ( 2+1 \right )^{2}+\left ( 5+2 \right )^{2}}
AC= \sqrt{9+49}= \sqrt{58}
Length \, of \, BD= \sqrt{\left ( -3-4 \right )^{2}+\left ( 0-3 \right )^{2}}
BD= \sqrt{49+9}= \sqrt{58}

AC = BD (Diagonals)
Hence, ABCD is a rectangle because
AB = CD, BC = DA, AC = BD

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