AP EAMCET · Maths · Pair of Lines
The three sides of a triangle are given by \(\left(x^2+7 x y+2 y^2\right)(y-1)=0\). Then the centroid of that triangle is
- A \(\left(\frac{2}{3}, 0\right)\)
- B \(\left(\frac{7}{3}, \frac{2}{3}\right)\)
- C \(\left(\frac{-7}{3}, \frac{2}{3}\right)\)
- D \(\left(\frac{1}{3}, \frac{4}{3}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{-7}{3}, \frac{2}{3}\right)\)
Step-by-step Solution
Detailed explanation
The three sides are \(y-1=0\) and \(x^2+7xy+2y^2=0\). The lines \(x^2+7xy+2y^2=0\) pass through the origin. Thus, one vertex is \( (0,0) \). The other two vertices are the intersections of \(y=1\) with \(x^2+7xy+2y^2=0\). Substitute \(y=1\) into \(x^2+7xy+2y^2=0\):…
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