If the coordinates of the vertex A of a square ABCD are (3, -2) and the equation of diagonal BD is 3x - 7y + 6 = 0, find the equation of the diagonal AC. Also find the coordinates of the centre of the square.

Diagonals AC and BD of the square ABCD bisect each other at right angle at O.

∴ O is the mid-point of AC and BD.

Equation of BD is 3x - 7y + 6 = 0

⇒ 7y = 3x + 6

⇒ y = $\dfrac{3}{7}x + \dfrac{6}{7}$.

∴ Slope of BD = m₁ = $\dfrac{3}{7}$

Let slope of AC be m₂. Since, BD and AC are perpendicular,

$\therefore m$1 \times m2 = -1 \\[1em] \Rightarrow \dfrac{3}{7}m2 = -1 \\[1em] \Rightarrow m2 = -\dfrac{7}{3}.

Equation of AC will be

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - (-2) = -\dfrac{7}{3}(x - 3) \\[1em] \Rightarrow 3(y + 2) = -7(x - 3) \\[1em] \Rightarrow 3y + 6 = -7x + 21 \\[1em] \Rightarrow 7x + 3y = 15

Now we will find the coordinates of O, the points of intersection of AC and BD

⇒ 3x - 7y = -6 ….(i)
⇒ 7x + 3y = 15 ….(ii)

Multiplying (i) by 3 and (ii) by 7, we get,

⇒ 9x - 21y = -18 ….(iii)
⇒ 49x + 21y = 105 …(iv)

Adding (iii) and (iv) we get,

⇒ 9x + 49x - 21y + 21y = -18 + 105

⇒ 58x = 87

⇒ x = $\dfrac{87}{58} = \dfrac{3}{2}.$

Putting value of x in (i) we get,

$\Rightarrow 3 \times \dfrac{3}{2} - 7y = -6 \\[1em] \Rightarrow \dfrac{9}{2} - 7y = -6 \\[1em] \Rightarrow \dfrac{9}{2} + 6 = 7y \\[1em] \Rightarrow \dfrac{9 + 12}{2} = 7y \\[1em] \Rightarrow \dfrac{21}{2} = 7y \\[1em] \Rightarrow y = \dfrac{3}{2}.$

Hence, the equation of AC is 7x + 3y - 15 = 0 and coordinates of the center are $\Big(\dfrac{3}{2}, \dfrac{3}{2}\Big)$.