OABC is a square, O is the origin and the points A and B are (3, 0) and (p, q). If OABC lies in the first quadrant, find the values of p and q. Also write down the equations of AB and BC.

The square OABC is plotted on the graph below:

$\text{OA} = \sqrt{(3 - 0)^2 + (0 - 0)^2} \\[1em] = \sqrt{3^2 + 0^2} \\[1em] = \sqrt{9} \\[1em] = 3. \\[1em] \text{AB} = \sqrt{(3 - p)^2 + (0 - q)^2} \\[1em] = \sqrt{(3 - p)^2 + q^2}$

Since, OA = AB (as sides of square are equal)

$\therefore \sqrt{(p - 3)^2 + q^2} = 3 \\[1em] \Rightarrow (p - 3)^2 + q^2 = 9 \\[1em] \Rightarrow p^2 + 9 - 6p + q^2 = 9 \\[1em] \Rightarrow p^2 + q^2 - 6p = 0 \space …..(\text{i})$

By pythagoras theorem, OB² = OA² + AB².

$\Rightarrow \Big(\sqrt{(p - 0)^2 + (q - 0)^2}\Big)^2 = 3^2 + \Big(\sqrt{(3 - p)^2 + q^2}\Big)^2 \\[1em] \Rightarrow p^2 + q^2 = 9 + (3 - p)^2 + q^2 \\[1em] \Rightarrow p^2 + q^2 = 9 + 9 + p^2 - 6p + q^2 \\[1em] \Rightarrow p^2 - p^2 + q^2 - q^2 + 6p = 18 \\[1em] \Rightarrow 6p = 18 \\[1em] \Rightarrow p = 3.$

Substituting value of p in (i),

$\Rightarrow 3^2 + q^2 - 6(3) = 0 \\[1em] \Rightarrow q^2 + 9 - 18 = 0 \\[1em] \Rightarrow q^2 - 9 = 0 \\[1em] \Rightarrow q^2 - (3)^2 = 0 \\[1em] \Rightarrow (q - 3)(q + 3) = 0 \\[1em] \Rightarrow q - 3 = 0 \text{ or } q + 3 = 0 \\[1em] \Rightarrow q = 3 \text{ or } q = -3 \\[1em] \Rightarrow q = 3, -3.$

But q = -3 is not possible as the square is in 1st quadrant and the coordinates are positive in 1st quadrant.

∴ p = 3 and q = 3.

AB is parallel to y-axis,

∴ Equation of AB will be x = 3 or x - 3 = 0.

BC is parallel to x-axis,

∴ Equation BC will be y = 3 or y - 3 = 0.

Hence, the value of p = 3 and q = 3. Equation of AB is x - 3 = 0 and BC is y - 3 = 0.