A ( − 2 , 0 ) , B ( 2 , 3 ) , C ( 5 , − 1 ) A(-2,0), B(2,3), C(5, -1) A ( − 2 , 0 ) , B ( 2 , 3 ) , C ( 5 , − 1 )
B A → = ⟨ − 2 − 2 , 0 − 3 ⟩ = ⟨ − 4 , − 3 ⟩ \overrightarrow{BA}=\langle-2-2, 0-3\rangle=\langle-4, -3\rangle B A = ⟨ − 2 − 2 , 0 − 3 ⟩ = ⟨ − 4 , − 3 ⟩
B C → = ⟨ 5 − 2 , − 1 − 3 ⟩ = ⟨ 3 , − 4 ⟩ \overrightarrow{BC}=\langle5-2, -1-3\rangle=\langle3, -4\rangle BC = ⟨ 5 − 2 , − 1 − 3 ⟩ = ⟨ 3 , − 4 ⟩
B A → ⋅ B C → = − 4 ( 3 ) + ( − 3 ) ( − 4 ) = 0 \overrightarrow{BA}\cdot\overrightarrow{BC}=-4(3)+(-3)(-4)=0 B A ⋅ BC = − 4 ( 3 ) + ( − 3 ) ( − 4 ) = 0 Then B A → ⊥ B C → , \overrightarrow{BA}\perp\overrightarrow{BC}, B A ⊥ BC , and we see that the triangle A B C ABC A BC with vertices A ( − 2 , 0 ) , B ( 2 , 3 ) , C ( 5 , − 1 ) A(-2,0), B(2,3), C(5, -1) A ( − 2 , 0 ) , B ( 2 , 3 ) , C ( 5 , − 1 ) is a right triangle.
∣ B A → ∣ = ( − 4 ) 2 + ( − 3 ) 2 = 5 |\overrightarrow{BA}|=\sqrt{(-4)^2+(-3)^2}=5 ∣ B A ∣ = ( − 4 ) 2 + ( − 3 ) 2 = 5
∣ B C → ∣ = ( 3 ) 2 + ( − 4 ) 2 = 5 |\overrightarrow{BC}|=\sqrt{(3)^2+(-4)^2}=5 ∣ BC ∣ = ( 3 ) 2 + ( − 4 ) 2 = 5
A r e a A B C = 1 2 ∣ B A → ∣ ∣ B A → ∣ Area_{ABC}=\dfrac{1}{2}|\overrightarrow{BA}||\overrightarrow{BA}| A re a A BC = 2 1 ∣ B A ∣∣ B A ∣
= 1 2 ( 5 ) ( 5 ) = 12.5 ( u n i t s 2 ) =\dfrac{1}{2}(5)(5)=12.5({units}^2) = 2 1 ( 5 ) ( 5 ) = 12.5 ( u ni t s 2 )
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