The centroid of the triangle formed by the lines \(6x^2+xy-2y^2=0\) and \(x+2y+3=0\) is
Show Hint
- \(\left(\frac{3}{10},-\frac{23}{10}\right)\)
- \(\left(\frac{3}{10},-\frac{13}{10}\right)\)
- \(\left(-\frac{3}{10},\frac{5}{4}\right)\)
- \(\left(-\frac{3}{5},\frac{5}{4}\right)\)
The Correct Option is A
Solution and Explanation
Step 1: Factorize.
\[ 6x^2+xy-2y^2=0 \] \[ (3x+2y)(2x-y)=0 \] Lines: \[ 3x+2y=0 \] \[ 2x-y=0 \] Third line: \[ x+2y+3=0 \]
Step 2: Find vertices.
Solving pairwise: \[ (3x+2y=0,\;2x-y=0)\Rightarrow (0,0) \] \[ (3x+2y=0,\;x+2y+3=0)\Rightarrow \left(-\frac{3}{2},\frac{9}{4}\right) \] \[ (2x-y=0,\;x+2y+3=0)\Rightarrow \left(-\frac{3}{5},-\frac{6}{5}\right) \]
Step 3: Apply centroid formula.
\[ \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right) \] \[ =\left(\frac{0-\frac32-\frac35}{3},\frac{0+\frac94-\frac65}{3}\right) \] \[ =\left(\frac{3}{10},-\frac{23}{10}\right) \]
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