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

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

\[\Rightarrow y - 2 = - 5\left( x - 1 \right) \text { and } y - 2 = \frac{1}{5}\left( x - 1 \right)\]

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

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).