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Question
Show that A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are the vertices of a
rhombus ABCD.
Solution
A(-4, -7), B(-1, 2), C(8, 5), D(5, -4)
According to distance formula,
`AB=sqrt([-1-(-4)]^2 +[2-(-7)]^2`
`∴ AB = sqrt(3^2+9^2)`
`∴ AB =sqrt(9+81) `
`∴ AB = sqrt90` ...... (1)
`BC =sqrt([8-(-1)]^2+(5-2)^2)`
`∴ BC =sqrt(9^2+3^2)`
`∴ BC = sqrt(81+9)`
`∴ BC = sqrt90` ...... (2)
`CD =sqrt((5-8)^2 +(-4-5)^2)`
`∴ CD =sqrt((-3)^2 +(-9)^2)`
`∴ CD =sqrt(9+81)`
`∴ CD = sqrt90 ` ....... (3)
`AD = sqrt([5-(-4)]^2+ [-4-(-7)]^2)`
`∴ AD =sqrt(9^2+3^2)`
`∴ AD =sqrt(81+9)`
`∴ AD = sqrt90` ....... (4)
From (1), (2), (3) and (4)
AB = BC = CD = AD
∴` square `ABCD is a rhombus.
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