Question

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

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

Answer 1

Answer: (C)

$(-8,8)$

Let the formed triangle be $ABC$.
Family of straight lines is

$AB+\lambda AC\to 2x+3y-1+\lambda (x+2y-1)=0$

It represents altitude through $A$

So the given line passes through orthocentre and we get

$\lambda =-1$

So the given line $AO$ is $x+y=0$

$AO$ is perpendicular to $BC$

[Using the concept that if line $L_1$ and $L_2$ are at right angles to each other then

$m_1\times m_2=-1$

where $m_1$ and $m_2$ are slopes of $L_1$ and $L_2$ respectively ]
$(-1)\times \frac{-a}{b}=-1$
$\therefore a=-b$
$BA+\mu BC=0$ as it passes through $(0,0)$ we get $\mu =-1$
This line is perpendicular to $AC$

$\therefore -\frac{2-a}{3+a}\times (-\frac{1}{2})=-1$
$\Rightarrow 2-a=-6-2a$

$\therefore a=-8$ and $b=8$

Hence, option C.