prove that the area of a triangle with vertices (t,t-2),(t+2,t+2),(t+3,t) is independent of it.
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Heya User,
--> Directly Applying the formula for area of Δ -->
--> 2 Area [ Δ ] = | t ( t + 2 - t ) + ( t + 2 )( t - t + 2 ) + ( t + 3 )( t - 2 - t - 2 )|
=> 2 Area [ Δ ] = | 2t + 2t + 4 - 4t - 12 |
=> 2 Area [ Δ ] = | 4t - 4t - 8 | = | -8 | = 8
=> Area [ Δ ] = 8 / 2 = 4 ;
Hence, we see that the Area of Δ = 4 which is a constant .. independent of the cruel 't' :p
Well, I cud've introduced the formula but the subscripts are =_= ..
If uh need it, just ask me.. ATB for ur assignment..
--> Directly Applying the formula for area of Δ -->
--> 2 Area [ Δ ] = | t ( t + 2 - t ) + ( t + 2 )( t - t + 2 ) + ( t + 3 )( t - 2 - t - 2 )|
=> 2 Area [ Δ ] = | 2t + 2t + 4 - 4t - 12 |
=> 2 Area [ Δ ] = | 4t - 4t - 8 | = | -8 | = 8
=> Area [ Δ ] = 8 / 2 = 4 ;
Hence, we see that the Area of Δ = 4 which is a constant .. independent of the cruel 't' :p
Well, I cud've introduced the formula but the subscripts are =_= ..
If uh need it, just ask me.. ATB for ur assignment..
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