The equations of two altitudes of an equilate | vedclass

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(B) Let the two given altitudes be $L_1: \sqrt{3}x - y + 8 - 4\sqrt{3} = 0$ and $L_2: \sqrt{3}x + y - 12 - 4\sqrt{3} = 0$. The intersection of these t

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$y = 10$

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(B) Let the two given altitudes be $L_1: \sqrt{3}x - y + 8 - 4\sqrt{3} = 0$ and $L_2: \sqrt{3}x + y - 12 - 4\sqrt{3} = 0$. The intersection of these two altitudes is the orthocenter of the triangle. Adding $L_1$ and $L_2$: $(\sqrt{3}x - y + 8 - 4\sqrt{3}) + (\sqrt{3}x + y - 12 - 4\sqrt{3}) = 0$. $2\sqrt{3}x - 4 - 8\sqrt{3} = 0$ $\Rightarrow 2\sqrt{3}x = 4 + 8\sqrt{3}$ $\Rightarrow x = \frac{2 + 4\sqrt{3}}{\sqrt{3}} = 4 + \frac{2}{\sqrt{3}}$. Subtracting $L_1$ from $L_2$: $(\sqrt{3}x + y - 12 - 4\sqrt{3}) - (\sqrt{3}x - y + 8 - 4\sqrt{3}) = 0$. $2y - 20 = 0 \Rightarrow y = 10$. The orthocenter is $(4 + \frac{2}{\sqrt{3}}, 10)$. In an equilateral triangle,the orthocenter is the same as the centroid. Since the third altitude must pass through the orthocenter $(x_0, y_0) = (4 + \frac{2}{\sqrt{3}}, 10)$,we check the given options. Substituting $(4 + \frac{2}{\sqrt{3}}, 10)$ into $y = 10$,we see it satisfies the equation. Thus,the equation of the third altitude is $y = 10$.

The equations of two altitudes of an equilateral triangle are $\sqrt{3}x - y + 8 - 4\sqrt{3} = 0$ and $\sqrt{3}x + y - 12 - 4\sqrt{3} = 0$. The equation of the third altitude is

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