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

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

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

$$\Rightarrow 5x+y-7=0\text{ and }x-5y+9=0$$

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