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What is the area of the parallelogram with the given vertices? A(-1, 3), B(0, 4), C(2, 2), D(1, 1)

1 Answer

Alan P. · Stefan V.

Jan 8, 2016

${Area}_{ABCD} = 4$ $"Area"_("ABCD")=4$

Explanation:

${Slope}_{AB} = \frac{4 - 3}{0 - (- 1)} = 1$ $"Slope"_("AB") = (4-3)/(0-(-1)) = 1$

${Slope}_{AD} = \frac{1 - 3}{1 - (- 1)} = - 1$ $"Slope"_("AD")=(1-3)/(1-(-1))= -1$

Since

$XXX {Slope}_{AB} = - \frac{1}{{Slope}_{AD}}$ $color(white)("XXX")"Slope"_text(AB) = - 1/("Slope"_text(AD))$

$A B$ $AB$ and $A D$ $AD$ are perpendicular and the parallelogram is a rectangle.

Therefore

$X {Area}_{ABCD} = | A B | \times | A D |$ $color(white)("X")"Area"_("ABCD") = |AB| xx |AD|$

$XXXXXXX = \sqrt{{(4 - 3)}^{2} + {(0 - (- 1))}^{2}} \times \sqrt{{(1 - 3)}^{2} + {(1 - (- 1))}^{2}}$ $color(white)("XXXXXXX")=sqrt((4-3)^2+(0-(-1))^2)xxsqrt((1-3)^2+(1-(-1))^2)$

$XXXXXXX = \sqrt{2} \times 2 \sqrt{2}$ $color(white)("XXXXXXX")=sqrt(2)xx2sqrt(2)$

$XXXXXXX = 4$ $color(white)("XXXXXXX")=4$

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