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Question
Show that A(4, –1), B(6, 0), C(7, –2) and D(5, –3) are vertices of a square.
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Solution
The given points are A(4, –1), B(6, 0), C(7, –2) and D(5, –3).
AB = \[\sqrt{\left( 6 - 4 \right)^2 + \left( 0 + 1 \right)^2} = \sqrt{4 + 1} = \sqrt{5}\]
BC =\[\sqrt{\left( 6 - 7 \right)^2 + \left( 0 + 2 \right)^2} = \sqrt{1 + 4} = \sqrt{5}\]
CD =\[\sqrt{\left( 7 - 5 \right)^2 + \left( - 2 + 3 \right)^2} = \sqrt{4 + 1} = \sqrt{5}\]
AD = \[\sqrt{\left( 5 - 4 \right)^2 + \left( - 3 + 1 \right)^2} = \sqrt{1 + 4} = \sqrt{5}\]
AB = BC = CD = DA
Slope of AB = \[\frac{0 + 1}{6 - 4} = \frac{1}{2}\]
Slope of BC = \[\frac{- 2 - 0}{7 - 6} = - 2\]
Slope of CD = \[\frac{- 3 + 2}{5 - 7} = \frac{1}{2}\]
Slope of AD = \[\frac{- 3 + 1}{5 - 4} = - 2\]
Thus, AB perpendicular to BC and AD. Also, CD perpendicular to AD and AB.
So, all the sides are equal to each other and are perpendicular.
Thus, they form a square.
