Question

prove that the points(3,-2)(4,0)(6,-3) and (5,-5) are the vertices of the parallelogram​

Answer 1

Answer:

Let P(x,y) be the point of intersection of diagonals AC and 80 of ABCD.

Here mid-point of AC – Mid - point of BD i.e, diagonals AC and BD bisect each other.

We know that diagonals of a parallelogram bisect each other

Therefore ABCD is a parallelogram

Answer 2

Answer:

Hy mate here is your answer

Let the vertices of a parallelogram be A(−3,2), B(4,0), C(6,−3) and D(5,−5)

So by distance formula we have,

Distance between two points = (x2−x1)2+(y2−y1)2

AB=(4+3)2+(0+2)2=5CD=(5−6)2+(−5+3)2=5BC=(6−4)2+(−3)2=13AD=(5−3)2+(−5+2)2=13

Opposite site of a parallelogram are equal.

Therefore AB=CD=5 and BC=AD=13

ABCD is a parallelogram (PROVED)

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