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Q.
The joint equation of two altitudes of an equilateral triangle is
$(\sqrt{3} x-y+8-4 \sqrt{3})(-\sqrt{3} x-y+12+4 \sqrt{3})=0$
The third altitude has the equation
Straight Lines
Solution:
Clearly, the third altitude will be the bisector of obtuse angle between the given altitudes.
Now given two altitudes are
$ \sqrt{3} x-y+8-4 \sqrt{3}=0 $ .... (i)
and $ -\sqrt{3} x-y+12+4 \sqrt{3}=0$ .....(ii)
$ a_{1} a_{2}+b_{1} b_{2}=-\sqrt{3} \sqrt{3}+(-1)(-1)=-2<0$
$\therefore $ Obtuse angle bisector is:
$\frac{\sqrt{3} x-y+8-4 \sqrt{3}}{3+1}=-\frac{-\sqrt{3} x-y+12+4 \sqrt{3}}{3+1}$
or $ y=10$