Question

Find the coordinates of the vertices of a triangle, the equations of whose sides are

y (t₁ + t₂) = 2x + 2a t₁t₂, y (t₂ + t₃) = 2x + 2a t₂t₃ and, y (t₃ + t₁) = 2x + 2a t₁t₃.

Answer 1

y (t₁ + t₂) = 2x + 2a t₁t₂, y (t₂ + t₃) = 2x + 2a t₂t₃ and y (t₃ + t₁) = 2x + 2a t₁t₃
2x − y (t₁ + t₂) + 2a t₁t₂ = 0     ... (1)
2x − y (t₂ + t₃) + 2a t₂t₃ = 0     ... (2)
2x − y (t₃ + t₁) + 2a t₁t₃ = 0     ... (3)
Solving (1) and (2) using cross-multiplication method:

\[\frac{x}{- 2a t_2 t_3 \left( t_1 + t_2 \right) + 2a t_1 t_2 \left( t_2 + t_3 \right)} = \frac{y}{4a t_1 t_2 - 4a t_2 t_3} = \frac{1}{- 2\left( t_2 + t_3 \right) + 2\left( t_1 + t_2 \right)}\]

\[ \Rightarrow \frac{x}{2a {t_2}^2 \left( t_1 - t_3 \right)} = \frac{y}{4a t_2 \left( t_1 - t_3 \right)} = \frac{1}{2\left( t_1 - t_3 \right)}\]

\[ \Rightarrow x = a {t_2}^2 , y = 2a t_2\]

Solving (1) and (3) using cross-multiplication method:

\[\frac{x}{- 2a t_1 t_3 \left( t_1 + t_2 \right) + 2a t_1 t_2 \left( t_3 + t_1 \right)} = \frac{y}{4a t_1 t_2 - 4a t_1 t_3} = \frac{1}{- 2\left( t_3 + t_1 \right) + 2\left( t_1 + t_2 \right)}\]

\[ \Rightarrow \frac{x}{2a {t_1}^2 \left( t_2 - t_3 \right)} = \frac{y}{4a t_1 \left( t_2 - t_3 \right)} = \frac{1}{2\left( t_2 - t_3 \right)}\]

\[ \Rightarrow x = a {t_1}^2 , y = 2a t_1\]

Similarly, solving (2) and (3) using cross-multiplication method:

\[\frac{x}{- 2a t_1 t_3 \left( t_2 + t_3 \right) + 2a t_2 t_3 \left( t_3 + t_1 \right)} = \frac{y}{4a t_2 t_3 - 4a t_1 t_3} = \frac{1}{- 2\left( t_3 + t_1 \right) + 2\left( t_2 + t_3 \right)}\]

\[ \Rightarrow \frac{x}{2a {t_3}^2 \left( t_2 - t_1 \right)} = \frac{y}{4a t_3 \left( t_2 - t_1 \right)} = \frac{1}{2\left( t_2 - t_1 \right)}\]

\[ \Rightarrow x = a {t_3}^2 , y = 2a t_3\]

Hence, the coordinates of the vertices of the triangle are \[\left( a {t_1}^2 , 2a t_1 \right)\],

\[\left( a {t_2}^2 , 2a t_2 \right)\] and \[\left( a {t_3}^2 , 2a t_3 \right)\].