JEE Main 2024 (Online) 31st January Evening ShiftQuestion 1
Let $A(a,b)$, $B(3,4)$ and $C(-6,-8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $P(2a+3,7b+5)$ from the line $2x+3y-4=0$ measured parallel to the line $x-2y-1=0$ is
Points and Straight Lines
Solution
Centroid divides orthocentre-circumcentre: $G=\frac{2O+H}{3}$ with $O=(3,4),H=(-6,-8)$: $G=(0,0)$, so $a=b=0$. $P=(3,5)$. Along direction of $x-2y-1=0$, i.e. $(2,1)/\sqrt5$: point $(3+2t,5+t)$ on $2x+3y-4=0$: $6+4t+15+3t-4=0\Rightarrow t=-\frac{17}{7}$. Distance $=|t|\sqrt5=\frac{17\sqrt5}{7}$. Answer (B).