Let the given points be: P(-3, -14) = (x\(_1\), y\(_1\)) Q(t, -5) = (x\(_2\), y\(_2\)) PQ = 9 units (given) Using the distance formula, d = √ [ \((x_2 - x_1)^2 + (y_2 - y_1)^2\)] PQ = √ [ \((t - (-3))^2 + (-5 + 14)^2\)] \(\implies\) √ [ \((t + 3)^2 + 81\)] = 9 Squaring on both sides, ⇒ (t + 3)\(^2\) + 81 = 81 ⇒ (t + 3)\(^2\) = 0 ⇒ t + 3 = 0 ∴ t = -3