The vertices of a triangle are `[at_(1)t_(2),a(t_(1)+t_(2))]`,`[at_(2)t_(3),a(t_(2)+t_(3))]`, `[at_(3)t_(1),a(t_(3)+t_(1))]`. Find the orthocentre of the triangle.

The vertices of a triangle are `[at_(1)t_(2),a(t_(1)+t_(2))]`,`[at_(2)

1.	The vertices of a triangle are `[at_(1)t_(2),a(t_(1)+t_(2))]`,`[at_(2)t_(3),a(t_(2)+t_(3))]`, `[at_(3)t_(1),a(t_(3)+t_(1))]`. Find the orthocentre of the triangle.
Answer» Let `ABC` be a triangle whose vertices are `A[at_(1)t_(2),a(t_(1)+t_(2))]`, `B[at_(2)t_(3),a(t_(2)+t_(3))]` and `C[at_(1)t_(3),a(t_(1)+t_(3))]`. Then, Slope of `BC=(a(t_(2)+t_(3))-a(t_(1)+t_(3)))/(at_(2)t_(3)-at_(1)t_(3))=(1)/(t_(3))` Slope of `AC=(a(t_(1)+t_(3))-a(t_(1)+t_(2)))/(at_(1)t_(3)-at_(1)t_(2))=(1)/(t_(1))` So, the equation of a line through A perpendicular to `BC` is `y-a(t_(1)+t_(2))=-t_(3)(x-at_(1)t_(2))`........`(i)` and the equation of a line through `B` perpendicular to `AC` is `y-a(t_(2)+t_(3))=-t_(1)(x-at_(2)t_(3))`.........`(ii)` The point of intersection of Eqs. `(i)` and `(ii)` , is the orthocentre. On subtracting Eq. `(ii)` from Eq. `(i)`, we get `x=-a`. On putting `x=-a` in Eq. `(i)` , we get `y=a(t_(1)+t_(2)+t_(3)+t_(1)t_(2)t_(3))` Hence, the coordinates of the orthocentree are `[-a,a(t_(1)+t_(2)+t_(3)+t_(1)t_(2)t_(3))]`.

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The vertices of a triangle are `[at_(1)t_(2),a(t_(1)+t_(2))]`,`[at_(2)t_(3),a(t_(2)+t_(3))]`, `[at_(3)t_(1),a(t_(3)+t_(1))]`. Find the orthocentre of the triangle.

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