Question

The orthocentre of the triangle formed by (0, 0), (8, 0), (4 6) is.

• A. $\left( 4,\frac{8}{3} \right)$
• B. (3, 4)
• C. (4, 3)
• D. (–3, 4)

Answer 1

Answer: (A)

$\left( 4,\frac{8}{3} \right)$

Answer 2

Let the vertices of the triangle be O(0, 0) $A ( 8,0 )$ $B ( 4,6 )$ . The equation of an altitude through O and perpendicular to AB is $y = \frac { 2 } { 3 } x$ and the equation of an altitude through $A ( 8,0 )$ and perpendicular to OB is

$3 y = - 2 x + 16$. The two altitudes intersect at $\left( 4 , \frac { 8 } { 3 } \right)$.