Question

The extremities of the diagonal of a square are the points (1,5) and (8 , 8). Find the equation to its sides and the co-ordinates of the other vertices.

Answer 1

Slope of AC = $\frac{8-5}{8-1}=\frac{3}{7}.$
The lines AB and AD pass through (8, 8) and are inclined at an angle of $\pm {45}^{\circ }$ to AC whose slope is 3/7.
Hence proceeding as in Ex. 18 their equation are
AB = 5x - 2y - 24 = 0.
AD = 2x + 5y - 56 = 0.
CD is parallel to AB and passes through (1, 5);
$\therefore$ CD is 5x - 2y + 5 = 0.
CB is parallel to AD and passes through (1, 5);
$\therefore$ CB is 2x + 5y - 27 = 0.
Solving AB and CB, we get point B as
$\frac{x}{54+120}=\frac{y}{48+135}=\frac{1}{25+4}$
$\therefore Bis(\frac{174}{29},\frac{87}{29})$ or (6, 3).
Solving AD and CD, we get the Point D as
$\frac{x}{112-25}=\frac{y}{10+280}=\frac{1}{25+4}$
$\therefore Dis(\frac{87}{29},\frac{290}{29})$ or (3, 10).