Question

show that A(1,2) B(5,4) C(3,8) and D(-1,6) are the vertices of a
square
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Answer 1

Hello !!

If all distances AB, BC , CD and AD are equal. Are vertices of a square.

Find dAB.

dAB = √[(xB - xA)^2 + (yB - yA)^2]

dAB = √[(5 - 1)^2 + (4 - 2)^2]

dAB = √[(4)^2 + (2)^2]

dAB = √[16 + 4]

dAB = √20

Find dBC.

dBC = √[(xC - xB)^2 + (yC - yB)^2]

dBC = √[(3 - 5)^2 + (8 - 4)^2]

dBC = √[(-2)^2 + (4)^2]

dBC = √[4 + 16]

dBC = √20

Find dCD.

dCD = √[(xD - xC)^2 + (yD - yC)^2]

dCD = √[(-1 - 3)^2 + (6 - 8)^2]

dCD = √[(-4)^2 + (2)^2]

dCD = √[16 + 4]

dCD = √20

Find dAD.

dAD = √[(xD - xA)^2 + (yD - yA)^2]

dAD = √[(-1 - 1)^2 + (6 - 2)^2]

dAD = √[(-2)^2 + (4)^2]

dAD = √[4 + 16]

dAD = √20

With this informations, we have.

AB = BC = CD = AD = L

Final result : yeah, are the vertices of a square.

I hope I have collaborated !