MHT CET · Maths · Pair of Lines
The centroid of the triangle formed by the lines \(x+3 y=10\) and \(6 x^2+x y-y^2=0\) is
- A \(\left(\frac{1}{3}, \frac{-7}{3}\right)\)
- B \(\left(\frac{-1}{3}, \frac{-7}{3}\right)\)
- C \(\left(\frac{-1}{3}, \frac{7}{3}\right)\)
- D \(\left(\frac{1}{3}, \frac{7}{3}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{-1}{3}, \frac{7}{3}\right)\)
Step-by-step Solution
Detailed explanation
Lines represented by the equation \(6 x^2+x y-y^2=0\) are \(y=3 x\) and \(y=-2 x\) The co-ordinates of the vertices of the triangle formed by above lines with \(x+3 y=10\) are \((0,0),(1,3)\) and \((-2,4)\).
\(\therefore \quad \text { Centroid }=\left(\frac{0+1-2}{3}, \frac{0+3+4}{3}\right)=\left(\frac{-1}{3}, \frac{7}{3}\right)\)
\(\therefore \quad \text { Centroid }=\left(\frac{0+1-2}{3}, \frac{0+3+4}{3}\right)=\left(\frac{-1}{3}, \frac{7}{3}\right)\)
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