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. $(-4,4)$
• D. $(0,7)$

Answer 1

Answer: (C)

$(-4,4)$

Slope of $OA$ is $-1$

$\therefore$ slope of $ax+by=1$ is 1

$\therefore \frac{-a}{b}=1$

$a=-b$ ----(1)

$\therefore B(\frac{3-2b}{2(3a-2b)},\frac{a-1}{3a-2b})$

$\therefore$ line perpendicular to $x+2y=1$ passing through origin

B line on $y=2x$

So, $\frac{a-1}{3a-2b}=2\left(\frac{3-2b}{2(3a-2b)}\right)$

$3a^2-3a-2ab+2b=9a-6ab-6b+4b^2$

From 1 $3a^2-3a+2a^2-2a=9a+6a^2+6a+4a^2$

$4a^2+20a=0$

$a=0,a=-4$

$(a,b)=(-4,4)$ $b=4$