JEE Main 2024 (Online) 8th April Morning ShiftQuestion 1
If the orthocentre of the triangle formed by the lines $2x+3y-1=0$, $x+2y-1=0$ and $ax+by-1=0$, is the centroid of another triangle, whose circumcentre and orthocentre respectively are $(3,4)$ and $(-6,-8)$, then the value of $|a-b|$ is $\_\_\_\_\_\_$ .
Points and Straight Lines
Answer$16$
Solution
Centroid $G$ of the other triangle: $H=3G-2O\Rightarrow G=\frac{H+2O}{3}=\frac{(-6,-8)+(6,8)}3=(0,0)$. So the orthocentre of the given triangle is the origin. Lines 1 and 2 meet at $(-1,1)$; the altitude from this vertex passes through $O$ and is perpendicular to $ax+by=1$, so $(a,b)\parallel(-1,1)$: $b=-a$. Vertex $2x+3y=1\cap ax+by=1$ lies on the altitude through $O$ perpendicular to $x+2y=1$, i.e. $y=2x$: $(\frac18,\frac14)$. Then $\frac a8+\frac b4=1$, with $b=-a$: $a=-8,\ b=8$. $|a-b|=16$.