The vertices of a triangle are
$[a t_{1} t_{2}, a (t_{1} + t_{2})], [a t_{2} t_{3}, a (t_{2} + t_{3})], [a t_{3} t_{1}, a (t_{3} + t_{1})]$ . Find the coordinates of its orthocentre.

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Solution

Let the given points be $A, B$ and $C$ respectively.
Slope of $B C$ by $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ is $\frac{1}{t_{3}}$
Hence the line through $A$ perpendicular to $B C$ is
$y - a (t_{1} + t_{2}) = - t_{3} (x - a t_{1} t_{2}) . . . . (1)$
Similarly the line through $B ⊥$ to $C A$ is
$y - a (t_{2} + t_{3}) = - t_{3} (x - a t_{2} t_{3}) . . . (2)$ Subtracting, we get $x = - a$ and hence $y = a (t_{1} + t_{2} + t_{3} + t_{1} t_{2} t_{3})$

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The vertices of a triangle are[at1t2,a(t1+t2)],[at2t3,a(t2+t3)],[at3t1,a(t3+t1)]. Find the coordinates of its orthocentre.

Let the given points be A,B and C respectively.Slope of BC by y2−y1x2−x1 is 1t3Hence the line through A perpendicular to BC isy−a(t1+t2)=−t3(x−at1t2)....(1)Similarly the line through B⊥ to CA isy−a(t2+t3)=−t3(x−at2t3)...(2) Subtracting, we get x=−a and hence y=a(t1+t2+t3+t1t2t3)

The vertices of a triangle are
$[a t_{1} t_{2}, a (t_{1} + t_{2})], [a t_{2} t_{3}, a (t_{2} + t_{3})], [a t_{3} t_{1}, a (t_{3} + t_{1})]$ . Find the coordinates of its orthocentre.