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

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



The sides AB and AD pass through the point A(1,2) and make an angle of ${45}^{\circ }$ with AC whose slope is $\frac{3}{2}$.

Equations of AB and AD are given by

$y-2=\frac{{\displaystyle \frac{3}{2}}\pm \tan {45}^{\circ }}{1\mp {\displaystyle \frac{3}{2}}\tan {45}^{\circ }}\left(x-1\right)$

$\Rightarrow y-2=\frac{{\displaystyle 3}\pm 2}{{\displaystyle 2\mp 3}}\left(x-1\right)$

$$\begin{gathered} \Rightarrow y-2=-5\left(x-1\right)\mathrm{and}y-2=\frac{1}{5}\left(x-1\right) \\ \Rightarrow 5x+y-7=0\mathrm{and}x-5y+9=0 \end{gathered}$$

Thus, the equations of AB and AD are $5x+y-7=0\mathrm{and}x-5y+9=0$, respectively.

Since BC is parallel to AD, the equation of BC is $x-5y+\lambda =0$.

This line passes through C (5,8).

$5-40+\lambda =0\Rightarrow \lambda =35$

So, the equation of BC is $x-5y+35=0$.

Since CD is parallel to AB, the equation of CD is$5x+y+\lambda =0$.

This line passes through C (5, 8).

$25+8+\lambda =0\Rightarrow \lambda =-33$

So, the equation of CD is $5x+y-33=0$.

Solving equation of AB and BC, we get B as (0, 7).

Solving equation of AD and CD, we get D as (6, 3).


Hence, the other two vertices are (0, 7) and (6, 3).