Straight Lines
Orthocentre as Centroid — Finding |a−b|
nta_pyq_2024_apr
Grade 11
Question:
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 _____
Step-by-Step Solution
Key Concept: Centroid of second triangle: on Euler line, centroid divides circumcentre-orthocentre in $1:2$: $G=\left(\frac{3\cdot1+(-6)\cdot1}{1+2}\right)$... wait, $G=\frac{O+2C}{3}$? No, $G$ divides $OH$ in $1:2$ from $O$: $G=\left(\frac{2\cdot3+(-6)}{3},\frac{2\cdot4+(-8)}{3}\right)=(0,0)$.
Centroid of second triangle $=(0,0)$. Orthocentre of first triangle at $(0,0)$. Solving gives $a=-8,b=8$, $|a-b|=16$.
Correct Answer: 16