Question

Orthocenter of the triangle formed by joining the points$\left( 2,\frac{1}{2} \right)$; $\left( 3,\frac{1}{3} \right)$and$\left( 4,\frac{1}{4} \right)$; is-

• A. $\left( \frac{1}{24},24 \right)$
• B. $\left( - \frac{1}{24},24 \right)$
• C. $\left( - \frac{1}{24},–24 \right)$
• D. $\left( 24,\frac{1}{24} \right)$

Answer 1

Answer: (C)

$\left( - \frac{1}{24},–24 \right)$

Answer 2

Coordinates of the vertices are of the form $\left( i,\frac{1}{i} \right)$,

I = 1, 2, 3, 4.

These coordinates satisfy the equation of rectangular hyperbola xy = 1. The orthocenter of the triangle formed by three such coordinates$\left( a,\frac{1}{a} \right)$; $\left( b,\frac{1}{b} \right)$ and $\left( c,\frac{1}{c} \right)$also lies on the hyperbola and given by $\left( - \frac{1}{abc},–abc \right)$. That is $\left( - \frac{1}{2 \times 3 \times 4},–2 \times 3 \times 4 \right)$ =$\left( - \frac{1}{24},–24 \right)$.