(d) (0, 0)Let the vertices of Δ ABC be given as: A(0, 0), B(3, 0) and C(0, 4) The orthocentre O is the point of intersection of the altitudes drawn from the vertices of Δ ABC on the opposite sides. Slope of BC = \(rac{4-0}{0-3}\) = \(rac{-4}{3}\)∴ Slope of AD ⊥ BC = \(rac{4}{3}\) (Slope of BC x Slope of AD = –1)Equation of line through A(0, 0) having slope \(rac{4}{3}\) isy - 0 = \(rac{4}{3}\) (x -0) ⇒ 4y = 3x ...(i) Similarly, slope of AC = \(rac{4-0}{0-3}\) = ∞Slope of line BE ⊥ AC = \(rac{1}{\infty}\) = 0∴ Equation of BE : (y – 0) = 0 (x – 3) ⇒ y = 0 ...(ii) From (i), \(x\) = 0 Hence, the orthocentre is (0, 0).