TS EAMCET · Maths · Pair of Lines
If the sides of a triangle \(A B C\) are \(2 x^2-y^2=0\), \(x+y-1=0\) and the sides of another triangle \(P Q R\) are \(2 x^2-5 x y+2 y^2=0,7 x-2 y-12=0\), then the distance between the centroid of \(\triangle A B C\) and the orthocentre of \(\triangle P Q R\) is
- A \(\frac{4}{3} \sqrt{261}\)
- B \(\frac{1}{3} \sqrt{165}\)
- C \(2 \sqrt{29}\)
- D \(56 \sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{29}\)
Step-by-step Solution
Detailed explanation
For \(\triangle A B C\), we have sides are \(2 x^2-y^2=0, x+y-1=0\) \((\sqrt{2} x-y)(\sqrt{2} x+y)=0, x+y-1=0\) \(\sqrt{2} x-y=0, \sqrt{2} x+y=0, x+y-1=0\) On solving these, we get vertices \((0,0),(\sqrt{2}-1,2-\sqrt{2}),(-\sqrt{2}-1,2+\sqrt{2})\) For \(\triangle P Q R\), we…
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