JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Straight Lines

A(a,b), B(3,4), C(-6,-8) are the centroid, circumcentre and orthocentre respectively. On the Euler line, the centroid divides the segment joining the circumcentre and orthocentre in the ratio 1:2 (measured from the circumcentre).

So $G=\frac{2\cdot (\text{circumcentre})+1\cdot (\text{orthocentre})}{3}=\frac{2(3,4)+(-6,-8)}{3}=\frac{(6,8)+(-6,-8)}{3}=(0,0)$.

Hence $a=0,b=0$, so $P(2a+3,7b+5)=P(3,5)$.

To find the distance from $P(3,5)$ to the line $2x+3y-4=0$ measured parallel to $x-2y-1=0$: the line $x-2y-1=0$ has slope $\frac{1}{2}$, so its direction makes $\cos \theta =\frac{2}{\sqrt{5}},\sin \theta =\frac{1}{\sqrt{5}}$.

A general point on the line through $P$ in this direction is $\left(3+\frac{2r}{\sqrt{5}},5+\frac{r}{\sqrt{5}}\right)$. Substituting into $2x+3y-4=0$:

$2\left(3+\frac{2r}{\sqrt{5}}\right)+3\left(5+\frac{r}{\sqrt{5}}\right)-4=0$

$6+\frac{4r}{\sqrt{5}}+15+\frac{3r}{\sqrt{5}}-4=0$

$17+\frac{7r}{\sqrt{5}}=0\Rightarrow r=-\frac{17\sqrt{5}}{7}$.

The required distance is $|r|=\frac{17\sqrt{5}}{7}$.