If the orthocentre of the triangle formed by the line `2x+ 3y - 1 = 0 , x + 2y - 1 = 0 , ax + by - 1 = 0 ` is at origin , then find (a,b)

Options

A. (6,4)
B. `(-3,3)`
C. `(-8,8)`
D. `(0,7)`

Answer

C

Explanation

The equation of a line through A i.e., the point of intersection of AB and AC is .
`(x + 2y -1)+lambda(2x + 3y -1) =0" "…..(i)`
If it passes through (0,0) then `-1- lambda = 0 rArr lambda = -1`
Substituting `lambda = - 1` in (i), we get
x + y =0 as the equation of AD.
Since `AD bot BC`. Therefore,
`-1 xx -(a)/(b) = - 1 rArr a+ b = 0" "....(ii)`
ltbr Similarly, by applying the condition that BO is perpendicular to CA, we get
` a + 2b = 8" "......(iii)`
Sloving (ii) and (iii), we get a =- 8, b = 8 .