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
Sides of the triangle are: \(y-1=0 \implies y=1\) \(x^2+7xy+2y^2=0 \implies \left(x+\frac{7+\sqrt{41}}{2}y\right)\left(x+\frac{7-\sqrt{41}}{2}y\right)=0\) Lines are: \(L_1: y=1\), \(L_2: 2x+(7+\sqrt{41})y=0\), \(L_3: 2x+(7-\sqrt{41})y=0\) Vertices:…
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