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Find the area of a triangle whose vertices are$(at_1^2, 2at_1), (at_2^2, 2at_2)$ and $(at_3^2, 2at_3)$
Given:
Vertices of a triangle are $(at_1^2, 2at_1), (at_2^2, 2at_2)$ and $(at_3^2, 2at_3)$.
To do:
We have to find the area of the given triangle.
Solution:
Let $A(at_1^2, 2at_1), B(at_2^2, 2at_2)$ and $C(at_3^2, 2at_3)$ be the vertices of a $\triangle ABC$.
We know that,
Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by,
Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$
Therefore,
Area of triangle $ABC=\frac{1}{2}[a t_{1}^{2}(2 a t_{2}-2 a t_{3})+a t_{2}^{2}(2 a t_{3}-2 a t_{1})+a t_{3}^{2}(2 a t_{1}-2 a t_{2})]$
$=\frac{1}{2}[2 a^{2} t_{1}^{2} t_{2}-2 a^{2} t_{1}^{2} t_{3}+2 a^{2} t_{2}^{2} t_{3}-2 a^{2} t_{2}^{2}t{1}+2 a^{2} t_{3}^{2} t_{1}-2 a^{2} t_{3}^{2} t_{2}]$
$=\frac{1}{2} \times 2 a^{2}[t_{1}^{2} t_{2}-t_{1}^{2} t_{3}+t_{2}^{2} t_{3}-t_{2}^{2} t_{1}+t_{3}^{2} t_{1}-t_{3}^{2} t_{2}]$
$=a^{2}[t_{1}^{2}(t_{2}-t_{3})+t_{2}^{2}(t_{2}-t_{3})-t_{1}(t_{2}^{2}-t_{3}^{2})]$
$=a^{2}[t_{1}^{2}(t_{2}-t_{3})+t_{2} t_{3}(t_{2}-t_{3})-t_{1}(t_{2}+t_{3})(t_{2}-t_{3})]$
$=a^{2}(t_{2}-t_{3})[t_{1}^{2}+t_{2} t_{3}-t_{1} t_{2}-t_{1} t_{3}]$
$=a^{2}(t_{2}-t_{3})[t_{1}^{2}-t_{1} t_{2}-t_{1} t_{3}+t_{2} t_{3}]$
$=a^{2}(t_{2}-t_{3})[t_{1}(t_{1}-t_{2})-t_{3}(t_{1}-t_{2})]$
$=a^{2}(t_{2}-t_{3})(t_{1}-t_{2})(t_{1}-t_{3})$
$=a^{2}(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})$
The area of the given triangle is $a^{2}(t_{1}-t_{2})(t_{2}-t_{3})(t_{3}-t_{1})$ sq. units.
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