A triangle is formed by the lines $x-2y-6=0$ , $ 3x−y+6=0$, $7x+4y−24=0$.
Find the equation of the line that bisects the inner angle of the triangle that is facing the side $7x+4y−24=0$.
I tried to find the intersect point of three equations by put them equal two by two. However, I don't know what to do next. could someone help me, please?
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$\begingroup$HINT: if point $P(x,y)$ is in the bisector of angle between two lines $x-2y-6=0$ and $3x-y+6=0$, then the distances from this point to each lines are equal. So, you get two lines which are bisector. Check which is the inner bisector.
HINT 2: See this link for distance between point $P(x_0,y_0)$ and the line $ax+by+c=0$. Using this formula, you have distance from $P(x_0,y_0)$ to two lines. Solve the system of equations, you get $P$.
$\endgroup$ 1 $\begingroup$This has nothing to do with triangle. You have two lines $x-2y=6$ and $3x-y=6$, intersecting at the point $P=(6/5,-12/5)$, and want an equation for the line bisecting the angle, presumably with positive slope. My suggestion, perhaps not as good as that of @GAVD, is to take any other point $Q_1$ on the first line, find its distance $d$ from $P$, and then find a point $Q_2$ on the second line at a distance of $d$ from $P$. Then build the rhombus whose vertices are, in order, $Q_1,P,Q_2,P'$, and draw the bisector from $P$ to $P'$. If you’re doing it by hand computation, it’ll be a mess.
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