[Toán 10] Chứng minh

G

goodgirla1city

H

huongmot

Lấy A, B là 2 điểm \in \triangle

\Rightarrow axA+byA+c=axB+byB+c=0ax_A + by_A + c = ax_B +by_B + c =0

\Rightarrow a(xAxB)=b(yAyB) a(x_A - x_B) = - b(y_A-y_B)

\Rightarrow xAyB=b(yAyB)a x_A - y_B=\dfrac{-b(y_A-y_B)}{a}

Tọa độ BA=(xAxB;yAyB)=(b(yAyB)a;yAyB)\vec{BA} = (x_A - x_B; y_A - y_B) = (\dfrac{-b(y_A-y_B)}{a};y_A - y_B)

Xét BA.n=b(yAyB)a.a+(yAyB)b=0\vec{BA}.\vec{n} = \dfrac{-b(y_A-y_B)}{a}.a + (y_A - y_B)b = 0

\Rightarrow BAn(đpcm)\vec{BA}\bot\vec{n} (đpcm)

Ta có n.u=0\vec{n}.\vec{u} =0 \Rightarrow nu\vec{n}\bot\vec{u} \Rightarrow u//(đpcm)\vec{u} // \triangle (đpcm)
 
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