Question number 13

Ans is hypotenuse (t₁t₂ ,t₁t₂) ;(0, 0) fixed

let eqns of sides be (1+t1)x + t1*y + t1(1+t1)= 0

and (1+t2)x + t2*y + t2(1+t2)= 0.

obviously these intersect y= 0 or x axis at ( – t1, 0) and ( – t2, 0). these are 2 of the vertices of the triangle.

3^rd vertex can be found out by solving the above 2 eqns and u get it as (t1t2, – (1+t1)(1+t2)). as we have all 3 vertices, we find orthocentre by the std method (writing eqns of perpendiculars from any 2 vertices on the opposite sides and then solving them. here, obviously one of the perpendiculars would be x= t1t2 and the other perpendicular will have to be calculated).

orthocentre comes out to be (t1t2, t1t2).

since both the coordinates (h, k) or (x, y) of the orthocentre are equal, hence it obviously lies on the line x – y= 0. if any one of the t’s is 0, then t1t2 will be 0. in that case orthocentre will be (0, 0).