(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 = \(\frac{4-0}{0-3}\) = \(\frac{-4}{3}\)
∴ Slope of AD ⊥ BC = \(\frac{4}{3}\) (Slope of BC x Slope of AD = –1)
Equation of line through A(0, 0) having slope \(\frac{4}{3}\) is
y - 0 = \(\frac{4}{3}\) (x -0) ⇒ 4y = 3x ...(i)
Similarly, slope of AC = \(\frac{4-0}{0-3}\) = ∞
Slope of line BE ⊥ AC = \(\frac{1}{\infty}\) = 0
∴ Equation of BE : (y – 0) = 0 (x – 3) ⇒ y = 0 ...(ii)
From (i), \(x\) = 0
Hence, the orthocentre is (0, 0).
