Question

If two opposite vertices of a square are (1, 2) and (5, 8), find the coordinates of its other two vertices and the equations of its sides.

Answer 1

Let A(1, 2), C(5, 8), $B(x_1,y_1)$, $D(x_2,y_2)$

Slope of AC $=\frac{8-2}{5-1}=\frac{6}{4}=\frac{3}{2}$

Let m be the slope of a line making an angle ${45}^{\circ }$ with AC.

$\therefore tan{45}^{\circ }=\left|\frac{m_1-\frac{3}{2}}{1+m\times \frac{3}{2}}\right|$

$1=\frac{m-\frac{3}{2}}{1+\frac{3m}{2}}$

$1+\frac{3m}{2}=m-\frac{3}{2}$ or, $1+\frac{3m}{2}=-(m-\frac{3}{2})$

$\frac{3m}{2}-m=\frac{-3}{2}-1$ or, $1+\frac{3m}{2}=-m+\frac{3}{2}$

$\frac{1}{2}m=\frac{-5}{2}$ or, $\frac{3m}{2}+m=\frac{3}{2}-1$

$m=-5$ or, $\frac{5m}{2}=\frac{1}{2}$

$m=\frac{1}{5}$

Consider the following figure:

Hence, equation of AD.

$y-2=-5(x-1)$

$y-2=-5x+5$

$5x+y=7$

Equation of CD,

$y-8=\frac{1}{5}(x-5)$

$y-8=\frac{x}{5}-1\Rightarrow y-8=\frac{x-5}{5}$

$=5y-40=x-5\Rightarrow 5y-x=35$

$\Rightarrow x-5y+35=0$

Hence co-ordinates are (6, 3), (0, 7)

The equation of AB is

$y-2=\frac{1}{5}(x-1)$

$\Rightarrow 5y-10=x-1$

$x-5y+9=0$

And the equation of BC is

$y-8=-5(x-5)$

$\Rightarrow y-8=-5x+25$

$5x+y-33=0$