Show that the following points are the vertices of a rectangle
A (0,-4), B(6,2), C(3,5) and D(-3,-1)
Solution
The given points are A (0,-4), B(6,2), C(3,5) and D(-3,-1).
`AB = sqrt((6-0)^2 +{2-(-4)}^2) = sqrt((6)^2 +(6)^2) = sqrt(36+36) = sqrt(72) = 6 sqrt(2) units`
`BC = sqrt(( 3-6)^2 + (5-2)^2) = sqrt((-3)^2 +(3)^2) = sqrt(9+9) = sqrt(18) = 3 sqrt(2) units`
` CD = sqrt((-3-3)^2 +(-1-5)^2) = sqrt((-6)^2 +(-6)^2) = sqrt(36+36) = sqrt(72) = 6 sqrt(2) units`
` AD = sqrt((-3-0)^2 + { -1-(-4)}^2) = sqrt((-3)^2 +(3)^2) = sqrt(9+9) = sqrt(18) = 3 sqrt(2) units`
` Thus , AB =CD = sqrt(10) " units and " BC = AD = sqrt(5) units`
`Also , AC = sqrt((3-0)^2 +{ 5-(-4)}^2) = sqrt((3)^2 +(9)^2 )= sqrt(9+81) = sqrt(90) = 3 sqrt(10) units`
`BD = sqrt((-3-6)^2 +(-1-2)^2) = sqrt((-9)^2 +(-3)^2) = sqrt(81+9) = sqrt(90) = 3 sqrt(10) units`
Also, diagonal AC = diagonal BD
Hence, the given points from a rectangle