How will we find the equation of AC?

Question

The vertex A of $\triangle$ABC is $(3,-1)$. The equations of median BE and angle bisector CF are $x-4y+10=0$ and $6x+10y-59=0$, respectively. What will be the equation of AC?

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I tried using the reflective property as its given that CF is the angle bisector. That way we can find the image of A about CF which would obviously lie on BC. But that doesn't help in finding AC as we need either another point on AC or AC's slope.

Also, I can't figure out how to use the fact that BE is the median. Is there a better approach to this problem?

Answer 1

Consider coordinates of C (h,k). So we have equation $$6h+10k=59$$. Moreover coordinates of E can be found by section formula. Now E lies on $$x-4y+10=0$$.

Now we have two equations in $$(h,k)$$. We will get the value of coordinate of C and then we can compute the equation of AC.

Answer 2

Equation of $BE, x-4y+10=0 \implies x = 4y - 10$ so coordinates of point $E$ on the line can be written as $(4y_1-10, y_1)$.

Equation of $CF, 6x+10y-59=0 \implies x = \frac{59-10y}{6}$ so coordinates of point $C$ on line $CF$ can be written as $(\frac{59-10y_2}{6}, y_2)$.

Coordinates of $A$ is $(3, -1)$. As $E$ is midpoint of $AC$,

$y_1 = \frac{y_2-1}{2}$ and $4y_1 - 10 = \frac{3 + (59-10y)/6}{2}$

Can you solve both equations to find coordinates of $E$ and equation of line passing through $A$ and $E$?