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
- A \(y- 2ax\, = 0\)
- B \(y- (a^2 + 1)x\, = 0\)
- C \(y+ x\, = 0\)
- D \((a - 1)^2x - (a + 1)^2y\, = 0\)
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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