Straight Lines
Centroid, Circumcentre, Orthocentre — Distance Parallel to Line
nta_pyq_2024_jan
Grade 11
Question:
Let $A(a,b)$, $B(3,4)$ and $(-6,-8)$ 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
$\dfrac{15\sqrt{5}}{7}$
$\dfrac{17\sqrt{5}}{6}$
$\dfrac{17\sqrt{5}}{7}$
$\dfrac{\sqrt{5}}{17}$
Step-by-Step Solution
Key Concept: Centroid divides segment joining circumcentre and orthocentre in ratio $1:2$. $A=\frac{2B+C}{3}$... actually $H=3G-2O$ (Euler line). $a=\frac{B_x+O_x\cdot2}{3}$... $G$ divides $O$ to $H$ in $1:2$: $G=(2\times3+(-6))/3=0, (2\times4+(-8))/3=0$. So $a=0,b=0$. $P=(3,5)$. Distance from $2x+3y-4=0$ measured parallel to $x-2y-1=0$.
$a=0,b=0,P=(3,5)$. Distance $=\frac{17\sqrt5}{7}$.
Correct Answer: 3