Question

If the straight line $2x+3y-1=0$ $x+2y-1=0$ and
$ax+by-1=0$ form a triangle with origin as orthocentre, then $(a,b)$ is equal to

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

Answer 1

Answer: (C)

The equation of a line through $A$ i.e. the point of intersection of
$AB$ and $AC$ is $(x+2y-1)+\lambda (2x+3y-1)=0$
If it passes through $(0,0)$, then $-1-\lambda =0$
$-1-\lambda =0$
$\Rightarrow \lambda =-1$
On substituting $\lambda =-1$ in Eq. (i),
we get $x+y=0$ as the equation of $AD$. Since, $AD\perp BC$, therefore
$1\times -\frac{a}{b}=-1$
$\Rightarrow a+b=0$......(ii)
Similarly, by applying the condition that $BO$ is perpendicular to $CA$, we get $a+2b=8$
On solving Eqs. (ii) and (iii), we get
$a=-8,b=8$. ........(iii)
On solving Eqs. (ii) and (iii), we get $a=-8,b=8$.