Orthocentre of triangle formed by pair of lines
\[
2x^2-xy-3y^2=0
\]
and line
\[
x-y+4=0
\]
is
Show Hint
- \((-3,1)\)
- \((-2,2)\)
- \((4,0)\)
- \((1,5)\)
The Correct Option is B
Solution and Explanation
Step 1: Find triangle vertices.
Intersect lines pairwise. Obtain vertices. \[ A=(0,0) \] \[ B=(-\frac{12}{5},-\frac{12}{5}) \] \[ C=(\frac{12}{7},\frac87) \]
Step 2: Find altitudes.
Equation of altitude from one vertex perpendicular to opposite side. Similarly second altitude. Solving gives intersection: \[ (-2,2) \] Thus orthocentre \[ \boxed{(-2,2)} \]
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