The ends of a diagonal of a square have co-ordinates (-2, p) and (p, 2). Find p if the area of the square is 40 sq. units.

Given,

Ends of a diagonal of a square are (-2, p) and (p, 2).

Area of square = 40 sq. units

By formula,

Area of square = (side)²

∴ (side)² = 40

⇒ side = $\sqrt{40} = 2\sqrt{10}$ units.

By formula,

Diagonal of a square = $\sqrt{2}$ × side = $\sqrt{2} \times 2\sqrt{10} = 2\sqrt{20}$.

By distance formula,

$d = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] \Rightarrow 2\sqrt{20} = \sqrt{[p - (-2)]^2 + [2 - p]^2} \\[1em] \Rightarrow 2\sqrt{20} =\sqrt{[p + 2]^2 + [2 - p]^2} \\[1em] \Rightarrow 2\sqrt{20} = \sqrt{p^2 + 4 + 4p + 4 + p^2 - 4p} \\[1em] \Rightarrow 2\sqrt{20} = \sqrt{2p^2 + 8}

On squaring both sides,

$\Rightarrow 4 \times 20 = 2p^2 + 8 \\[1em] \Rightarrow 2p^2 = 80 - 8 \\[1em] \Rightarrow 2p^2 = 72 \\[1em] \Rightarrow p^2 = \dfrac{72}{2} \\[1em] \Rightarrow p^2 = 36 \\[1em] \Rightarrow p = \sqrt{36} \\[1em] \Rightarrow p = \pm 6.$

Hence, p = ±6.