What kind of shape has vertices $A (- 1, - 4), B (1, - 1), C (4, 1), D (2, - 2)$ $A(-1, -4), B(1, -1), C(4,1), D(2, -2)$ ?

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Answer 1 · 2018-06-10T14:45:18

Shape of the figure is a SQUARE

$A (- 1, - 4), B (1, - 1), C (4, 1), D (2, - 2)$ $A(-1,-4), B(1,-1), C(4,1), D(2,-2)$

Slope of $¯ ¯¯¯¯ ¯ A B = \frac{- 1 + 4}{1 + 1} = \frac{3}{2}$ $bar(AB) = (-1+4) / (1 + 1) = 3/2$

Slope of $¯ ¯¯¯¯ ¯ C D = \frac{- 2 - 1}{2 - 4} = \frac{3}{2}$ $bar (CD) = (-2-1) / (2-4) = 3/2$

Slope of $¯ ¯¯¯¯ ¯ B C = \frac{1 + 1}{4 - 3} = - \frac{2}{3}$ $bar(BC) = (1+1)/(4-3) = -2/3$

Slope of $¯ ¯¯¯¯ ¯ A D = \frac{- 2 + 4}{2 + 1} = - \frac{2}{3}$ $bar (AD) = (-2+4) / (2+1) = -2/3$

$- - \to A B$ $vec(AB)$ “parallel “ $- - \to C D, - - \to B C$ $vec(CD), vec(BC)$ “ parallel “ $- - \to A D$ $vec(AD)$

$- - \to A B$ $vec(AB)$ “ perpendicular “ $- - \to B C$ $vec(BC)$ , $- - \to C D$ $vec(CD)$ “ perpendicular “ vec(AD)#

$- - \to A B = \sqrt{{(1 + 1)}^{2} + {(- 1 + 4)}^{2}} = \sqrt{13}$ $vec(AB) = sqrt((1+1)^2 + (-1+4)^2) = sqrt13$

$- - \to B C = \sqrt{{(4 - 1)}^{2} + {(1 + 1)}^{2}} = \sqrt{13}$ $vec(BC) = sqrt((4-1)^2 + (1+1)^2) = sqrt13$

Hence $- - \to A B = - - \to B C$ $vec(AB) = vec(BC)$

Therefore the shape of the figure is a SQUARE

What kind of shape has vertices A(−1,−4),B(1,−1),C(4,1),D(2,−2)A(-1, -4), B(1, -1), C(4,1), D(2, -2)?