The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.

Let the co-ordinates of point Q be (10, y).

Distance between the given points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Distance between P(2, -3) and Q(10, y):

$⇒ \sqrt{(10 - 2)^2 + (y - (-3))^2} = 10\\[1em] ⇒ (10 - 2)^2 + (y - (-3))^2 = 100\\[1em] ⇒ 8^2 + (y + 3)^2 = 100\\[1em] ⇒ 64 + y^2 + 9 + 6y = 100\\[1em] ⇒ y^2 + 6y + 73 = 100\\[1em] ⇒ y^2 + 6y + 73 - 100 = 0\\[1em] ⇒ y^2 + 6y - 27 = 0\\[1em] ⇒ y^2 + 9y - 3y - 27 = 0\\[1em] ⇒ (y^2 + 9y) - (3y + 27) = 0\\[1em] ⇒ y(y + 9) - 3(y + 9) = 0\\[1em] ⇒ (y + 9)(y - 3) = 0\\[1em] ⇒ y = -9 \text{ or } 3$

Hence, the required co-ordinates of the point Q are (10, -9) and (10, 3).