MHT CET · Maths · Straight Lines
The Centroid of the triangle formed by the lines \(6 x^2+x y-y^2=0\) and \(x+3 y-10=0\) 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
\( 6 x^2+x y-y^2=0 \)
\( \Rightarrow(3 x-y)(2 x+y)=0 \)
\( \Rightarrow 3 x-y=0,2 x+y=0 \)
\( \Rightarrow L_1 \equiv 3 x-y=0, L_2 \equiv 2 x+y=0 \text { and } L_3 \equiv x~+\) \(3 y=10 \)
point of intersection of
\(L_1\) and \(L_2\) is \((0,0), L_2\) and \(L_3\) is \((-2,4), L_3\) and \(L_1\) is \((1,3)\)
\( \Rightarrow \text { centroid of triangle } \equiv\left(\frac{0-2+1}{3}, \frac{0+4+3}{3}\right)\)\( \equiv\left(\frac{-1}{3}, \frac{7}{3}\right)\)
\( \Rightarrow(3 x-y)(2 x+y)=0 \)
\( \Rightarrow 3 x-y=0,2 x+y=0 \)
\( \Rightarrow L_1 \equiv 3 x-y=0, L_2 \equiv 2 x+y=0 \text { and } L_3 \equiv x~+\) \(3 y=10 \)
point of intersection of
\(L_1\) and \(L_2\) is \((0,0), L_2\) and \(L_3\) is \((-2,4), L_3\) and \(L_1\) is \((1,3)\)
\( \Rightarrow \text { centroid of triangle } \equiv\left(\frac{0-2+1}{3}, \frac{0+4+3}{3}\right)\)\( \equiv\left(\frac{-1}{3}, \frac{7}{3}\right)\)
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