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JEE Mains · Maths · STD 11 - 9. straight line

The circumcentre of a triangle lies at the origin and its centroid is the mid point of the line segment joining the points \((a^2 + 1 , a^2 + 1 )\) and \((2a, - 2a)\), \(a \ne 0\). Then for any \(a\) , the orthocentre of this triangle lies on the line

  1. A \(y- 2ax\, = 0\)
  2. B \(y- (a^2 + 1)x\, = 0\)
  3. C \(y+ x\, = 0\)
  4. D \((a - 1)^2x - (a + 1)^2y\, = 0\)
Verified Solution

Answer & Solution

Correct Answer

(D) \((a - 1)^2x - (a + 1)^2y\, = 0\)

Step-by-step Solution

Detailed explanation

Circumcentre \(=(0,0)\) Centroid \( = \left( {\frac{{{{\left( {a + 1} \right)}^2}}}{2},\frac{{{{\left( {a - 1} \right)}^2}}}{2}} \right)\) We know the circumcenter \((O)\), Centroid \((G)\) and orthocentre \((H)\) of a triangle lie on the line joining the \(O\) and \(G\) Also,…
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