question_answer

Orthocentre of the triangle whose vertices are (0, 0) (2, -1) and (1, 3) is                     [ISM Dhanbad1970; IIT 1967, 74]

A) \[\left( \frac{4}{7},\,\frac{1}{7} \right)\]

B) \[\left( -\frac{4}{7},\,-\frac{1}{7} \right)\]

C) (-4, -1)

D) (4, 1)

Correct Answer: B

Solution :

Equation of line BC is \[4x+y=7\], then equation of line AD is \[x-4y+k=0\] but it passes through (0,0) hence \[k=0\] thus equation of \[AD=x-4y=0\] .....(i) Similarly the equation of \[AC=3x-y=0\]and BE is   \[x+3y+1=0\]  .....(ii) On solving (i) and (ii), the required orthocentre is \[\left( \frac{-4}{7},\frac{-1}{7} \right)\].