Question #350448

Show that the points (-2,0), (2,3) and (5,-1) are the vertices of a right triangle. And find its area

Expert's answer

A(−2,0),B(2,3),C(5,−1)A(-2,0), B(2,3), C(5, -1)

BA→=⟨−2−2,0−3⟩=⟨−4,−3⟩\overrightarrow{BA}=\langle-2-2, 0-3\rangle=\langle-4, -3\rangle

BC→=⟨5−2,−1−3⟩=⟨3,−4⟩\overrightarrow{BC}=\langle5-2, -1-3\rangle=\langle3, -4\rangle

BA→⋅BC→=−4(3)+(−3)(−4)=0\overrightarrow{BA}\cdot\overrightarrow{BC}=-4(3)+(-3)(-4)=0

Then BA→⊥BC→,\overrightarrow{BA}\perp\overrightarrow{BC}, and we see that the triangle ABCABC with vertices A(−2,0),B(2,3),C(5,−1)A(-2,0), B(2,3), C(5, -1) is a right triangle.


∣BA→∣=(−4)2+(−3)2=5|\overrightarrow{BA}|=\sqrt{(-4)^2+(-3)^2}=5

∣BC→∣=(3)2+(−4)2=5|\overrightarrow{BC}|=\sqrt{(3)^2+(-4)^2}=5

AreaABC=12∣BA→∣∣BA→∣Area_{ABC}=\dfrac{1}{2}|\overrightarrow{BA}||\overrightarrow{BA}|

=12(5)(5)=12.5(units2)=\dfrac{1}{2}(5)(5)=12.5({units}^2)


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