Question

find the area of a triangle whose vertices are (at1 ​2 ,2at1),(at2 2 ,2at2),(at3 2 ,2at3).

Answer 1

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃) :
(1/2)|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|
Here, (x₁,y₁)=(at₁²,2at₁), (x₂,y₂)=(at₂²,2at₂), (x₃,y₃)=(at₃²,2at₃)
∴, area
=(1/2)|at₁²(2at₂-2at₃)+at₂²(2at₃-2at₁)+at₃²(2at₁-2at₂)|
=(1/2)|2a²t₁²t₂-2a²t₁²t₃+2a²t₂²t₃-2a²t₂²t₁+2a²t₃²t₁-2a²t₃²t₂|
=(1/2)2a²|t₁²t₂-t₁²t₃+t₂²t₃-t₂²t₁+t₃²t₁-t₃²t₂|
=a²|t₁²t₂-t₁²t₃+t₂²t₃-t₂²t₁+t₃²t₁-t₃²t₂|
=a²|t₁²(t₂-t₃)+t₂t₃(t₂-t₃)-t₁(t₂²-t₃²)|
=a²|t₁²(t₂-t₃)+t₂t₃(t₂-t₃)-t₁(t₂+t₃)(t₂-t₃)|
=a²|(t₂-t₃)(t₁²+t₂t₃-t₁t₂-t₁t₃)|
=a²|(t₂-t₃){t₁(t₁-t₂)-t₃(t₁-t₂)}|
=a²|(t₂-t₃)(t₁-t₂)(t₁-t₃)| Ans.