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Question
Prove that the points A(0, −1), B(−2, 3), C(6, 7) and D(8, 3) are the vertices of a rectangle ABCD?
Solution
Let the points (0, −1), (−2, 3), (6, 7), and (8, 3) be representing the vertices A, B, C, and D of a quadrilateral respectively.
Distance between the two points (x1, y1) and (x2, y2) is given by`sqrt((x_2-x_1)^2+(y_2-Y_1)^2)`
`thereforeAB=sqrt((0-(-2))^2+(-1-3)^2)``=sqrt((2)^2+(-4)^2)``=sqrt(4+16)=sqrt20=2sqrt5`
`BC=sqrt((-2-6)^2+(3-7)^2)=sqrt((-8)^2+(-4)^2)=sqrt64+16=sqrt80=4sqrt5`
`CD=sqrt((6-8)^2+(7-3)^2)=sqrt((-2)^2+(4)^2)=sqrt(4+16)=sqrt20=2sqrt5`
`AD=sqrt((8-0)^2+(3-(-1)^2))=sqrt((8)^2+(-8)^2)=sqrt64+16=sqrt80=4sqrt5`
Diagonal AC=`sqrt((0-6)^2+(-1-7)^2)=sqrt(6^2+(-8)^2)=sqrt100=10`
Diagonal AC=`sqrt((-2-8)^2+(3-3)^2)=sqrt((-10)^2+0^2)=sqrt100=10`
It can be observed that opposite sides of this quadrilateral are of the same length and also, the diagonals are of the same length.
Therefore, the given points are the vertices of a rectangle.
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