JEE Main 2014 · 19 Apr · Shift 1 · Mathematics Q78

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MathematicsSingle correct+4 / -1Question 78 of 90

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 $(2a,-2a)$ , $a\ne0$ . Then for any a, the orthocentre of this triangle lies on the line :

y-2ax=0

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

y+x=0

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

Answer and worked solution

Official answer: D

Centroid $G=\left(\frac{(a+1)^2}2,\frac{(a-1)^2}2\right)$ . With the circumcentre at O, the Euler line gives $H=3G$ , which lies on the line through O and G:
$(a-1)^2x-(a+1)^2y=0.$

Topics:Distance and section formulae Equations of a line

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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 78