Question

Quadrilateral ABCD has vertices A(- 3, 1), B(4, 2) B(4, 2), C(9, - 3) , and D(2, - 4) . aProve that quadrilateral ABCD is a rhombus b )Prove that ABCD is not a square .

Answer 1

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Step-by-step explanation:

The vectors representing the diagonals are ...

AC = C - A = (9, -3) -(-3, 1) = (12, -4)

BD = D - B = (2, -4) -(4, 2) = (-2, -6)

The dot-product of these vectors is the sum of the products of corresponding coordinates:

(12)(-2) +(-4)(-6) = -24 +24 = 0

When the dot-product is zero, the vectors are perpendicular. These vectors are different lengths (BD = 1/2·AC), so they are not the diagonals of a square.

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The midpoints of the diagonals will be in the same place if the quadrilateral is a rhombus.

(A +C)/2 = (B +D)/2

We can simplify our effort a bit by multiplying by 2.

A+C = B+D

(9, -3) +(-3, 1) = (2, -4) +(4, 2)

(6, -2) = (6, -2) . . . . . . . . . . . . . . . . yes, the midpoints are the same point

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We have shown the diagonals of the quadrilateral are different lengths and perpendicular bisectors of each other, so the quadrilateral IS A RHOMBUS, NOT A SQUARE.