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

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Answer 1 · 2018-06-10T14:45:18

Shape of the figure is a SQUARE

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

Slope of $bar(AB) = (-1+4) / (1 + 1) = 3/2$

Slope of $bar (CD) = (-2-1) / (2-4) = 3/2$

Slope of $bar(BC) = (1+1)/(4-3) = -2/3$

Slope of $bar (AD) = (-2+4) / (2+1) = -2/3$

$vec(AB)$ “parallel “ $vec(CD), vec(BC)$ “ parallel “ $vec(AD)$

$vec(AB)$ “ perpendicular “ $vec(BC)$ , $vec(CD)$ “ perpendicular “ vec(AD)#

$vec(AB) = sqrt((1+1)^2 + (-1+4)^2) = sqrt13$

$vec(BC) = sqrt((4-1)^2 + (1+1)^2) = sqrt13$

Hence $vec(AB) = vec(BC)$

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)?