Question

Show that A(4, –1), B(6, 0), C(7, –2) and D(5, –3) are vertices of a square.

Answer 1

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.