Question

The equation of the altitudes AD, BE, CF of a triangle ABC are x + y = 0, x – 4y = 0 and 2x – y = 0 respectively. If

A ŗ (t, – t) where t varies, then the locus of the centroid of triangle ABC is-

• A. y = –5x
• B. y = x
• C. x = – 5y
• D. x = – y

Answer 1

Answer: (C)

x = – 5y

Answer 2

All the altitudes pass through (0, 0). Hence origin is the orthocentre A บ (t, – t)

Let B บ (4t₁, t₁) and C บ (t₂, 2t₂) satisfying BE and CF.

m_BE = $\frac{1}{4}$, m_CF = 2, m_AC = $\frac{2t_{2} + t}{t_{2} - t}$ & m_AB = $\frac{t_{1} + t}{4t_{1} - t}$

so $\frac { 1 } { 4 }$ $\left( \frac{2t_{2} + t}{t_{2} - t} \right)$= –1 & 2 $\left( \frac{t_{1} + t}{4t_{1} - t} \right)$= –1

 t₂ =$\frac{t}{2}$ and t₁ = – $\frac{t}{6}$

so C บ $\left( \frac{t}{2},t \right)$ and B บ $\left( - \frac{2}{3}t, - \frac{t}{6} \right)$

Let G(x₁, y₁) be centroid of DABC and 't' varies

so x₁ = $\frac { 1 } { 3 }$ $\left( t - \frac{2}{3}t + \frac{t}{2} \right)$= $\frac{5t}{18}$ and

y₁ = $\frac { 1 } { 3 }$ $\left( - t - \frac{t}{6} + t \right)$= –$\frac{t}{18}$

Hence –x₁ = 5y₁  x = –5y