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

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 $AC$.
