A line $4x+y-1=0$ through $A(2,-7)$ meets the line $BC$ whose equation is $3x-4y-1=0$ at point $B$. Find Equation of Line $AC$, such that $AB=AC$. A line $4x+y-1=0$ through $A(2,-7)$ meets the line $BC$ whose equation is $3x-4y-1=0$ at point $B$. Find Equation of Line $AC$, such that $AB=AC$.
My Approach:
I took slope of line $AC$ as $m$.
And since $AB=AC$ that means Line $BC$ is equally inclined to $AC, AB$.
Now I used the formula for angle between two lines $\tan \theta = \bigg|\dfrac{(m_{1}-m_{2})}{1+m_{1}m_{2}}\bigg|$.
Slope of $AB$ is $-4$, slope of $BC$ is $\dfrac{3}{4}$ and Slope of $AC$ is $m$.
So, $\bigg| \dfrac{4m-3}{3m+4}\bigg|=\bigg  |\dfrac {19}{8}\bigg |$.
Note: $\dfrac{19}{8}$ is $\tan$ of angle between $AB$ and $BC$ and $\bigg| \dfrac{4m-3}{3m+4}\bigg|$ is $\tan$ of angle between $AB$ and $AC$.
My Doubt: After solving above relation I am obtaining two values of $m$. And those values are $\dfrac{-110}{25}$ and $\dfrac{-52}{89}$.
According to me Both answer must be correct but my professor said only $\dfrac{-52}{89}$ is correct.
Similar Question Straight lines - equation of line
 A: My approach:

*

*Find the distance from A to B. I got $\sqrt{\frac{18513}{361}}$. Ignore the square root for now, since we’ll have another square root in the next step, and they cancel out.

*Write out an equation for the distance from A to C, and set that equal to the distance from A to B. (The y-value for C will be $\frac{3}{4}x - \frac{1}{4}$.) You do this because you want to set the length of AB equal to AC. I got $\frac{5}{19}$ and $\frac{-1987}{475}$. Remember, these are the x-values of C.

To do this yourself, solve for $(x-2)^2+(\frac{3}{4}x - \frac{1}{4}+7)^2=\frac{18513}{361}$.

*

*Find the slope of the line from A to C (for both solutions above). I got $-4$ and $\frac{-52}{89}$, so your reasoning is correct ($\frac{-110}{25}$ should be $-4$, you probably just made a small calculation error.)

Now, why did I use this approach? Try solving for the x-value of B. It’s $\frac{5}{19}$, which was one of the x-values we got in step 2. This means that one of the values you got for point C was actually just point B! Technically, you could say $AB=AC$ if B was the same point as C, but your professor probably doesn’t allow that.
A: Solve the system of equations corresponding to the two given lines to find the coordinates of their point of intersection, which is $B.$ Find the distance between $A$ and $B.$ We'll call this distance $r.$ Solve the system of equations $$3x-4y=1,(x-2)^2+(y+7)^2=r^2$$. There are two solutions. One solution is the point $B$; the other solution is the point $C$. Since you now have the coordinates of both $A$ and $C$, you can find the equation of the line $AC$.
