If (1,2) and (3, 8) are a pair of opposite vertices of a square, find the
equation of the sides and diagonals of the square.

Question

If (1,2) and (3, 8) are a pair of opposite vertices of a square,
find the
equation of the sides and diagonals of the square

ramandeep.singh · Accepted Answer

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

AB and AD pass through A(1, 2) and make an angle of 45⁰ with AC, whose slope is 3. Therefore, equations of AB and AD are given by

$y - 2 = \frac{3 \pm \tan 45^{0}}{1 \pm 3 \tan 45^{0}} (x - 3) \Rightarrow y - 2 = \frac{3 \pm 1}{1 \pm 3} (x - 3) \Rightarrow y - 2 = \frac{2}{4} (x - 3) a n d y - 2 = \frac{- 4}{2} (x - 3) \Rightarrow 2 x - 4 y + 2 = 0 o r 2 x + y - 8 = 0$

Therefore, equation of AB is 2x - 4y + 2 = 0 and equation of AD is 2x + y - 8 = 0.

BC || AD. Therefore, equation of line BC is 2x + y + k = 0.

This passes through C(3, 8).

∴2(3) + 8 + k = 0

⇒k = -14

Therefore, equation of line BC is 2x + y - 14 = 0.

Since B is the point of intersection of AB and BC, therefore, on solving the equations of AB and BC, we get

$x = \frac{27}{5}, y = \frac{16}{5}$

Thus, coordinates of B are $(\frac{27}{5}, \frac{16}{5})$ .

E is the mid point of AC.

$∴ E = (\frac{1 + 3}{2}, \frac{2 + 8}{2}) = (2, 5)$

$\Rightarrow (\frac{p + \frac{27}{5}}{2}, \frac{q + \frac{16}{5}}{2}) = (2, 5) ∴ p = \frac{- 7}{5}, q = \frac{34}{5}$

Thus, coordinates of D are $(\frac{- 7}{5}, \frac{34}{5})$

If (1,2) and (3, 8) are a pair of opposite vertices of a square, find theequation of the sides and diagonals of the square.

If (1,2) and (3, 8) are a pair of opposite vertices of a square, find the
equation of the sides and diagonals of the square.