Question #2132

Prove that the quadrilateral ABCD is a rectangle given A(-8, -2) B (-3,3) C (3,-3) D(-2,-8) .

Expert's answer

Let's find the slope from the equations of the lines (y=mx+ b) passing through the points:

AB: A(-8, -2) B (-3,3)& m& = 1

BC: B (-3,3) C (3,-3)& m = -1

CD: C (3,-3) D(-2,-8) m = 1

DA: D(-2,-8) A(-8, -2)& m =-1

Thus AB is parallel to CD, BC is parallel to DA, as their slopes are equal to each other respectively, AB is perpendicular to BC, and BC is perpendicular to CD, as m(AB) = -m(BC) nad m(BC) = -m (CD). This shows that ABDC is a rectangle.

AB: A(-8, -2) B (-3,3)& m& = 1

BC: B (-3,3) C (3,-3)& m = -1

CD: C (3,-3) D(-2,-8) m = 1

DA: D(-2,-8) A(-8, -2)& m =-1

Thus AB is parallel to CD, BC is parallel to DA, as their slopes are equal to each other respectively, AB is perpendicular to BC, and BC is perpendicular to CD, as m(AB) = -m(BC) nad m(BC) = -m (CD). This shows that ABDC is a rectangle.

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