Question

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)`

Answer 1

We know area of triangle formed by three points (x₁y₁) , (x₂y₂), and (x₃y₃)is given by `triangle=1/2 [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`

The vertices are given as `(at_1^2,2at1),(at_2^2,2at_2),(at_3^2,2at_3)` 

`triangle =1/2[at_1^2(2at_2-2at_3)+at_2^2(2at_3-2at_1)+at_3^2(2at_1-2at_2)]` 

`=1/2xx2a^2[(t_1^2t_2-t_1^2t_3)+(t_2^2t_3-t_2^2t_1)+(t_3^2t_1-t_3^2t_2)]`

`=a^2[(t_1^2t_2-t_2^2 t_1)+(t_2^2 t_3-t_1^2 t_3)+(t_3^2t_1-t_3^2 t_2)]`

`=a^2 [t_1t_2(t_1-t_2)+t_3 (t_2^2-t_1^2)+t_3^2 (t_1-t_2)]`

`=a^2[(t_1-t_2) {t_1t_2-t_3(t_2+t_1)+t_3^2)]` 

`=a^2[(t_1-t_2){t_1t_2-t_3t_2-t_3t_1+t_3^2}`

`=a^2 [(t_1-t_2){t_2(t_1-t_3)-t_3 (-t_3+t_1)}]`

`=a^2[(t_1-t_2) (t_1-t_3)(t_2-t_3)]`

or,`triangle =a_2 (t_1-t_2) (t_2-t_3)(t_3-t_1)` assuming t₁> t₂, t₂ > t₃, t₃ > t₁