Equation of straight lines from a point making equal angles with given two lines. Given two lines $3x = 4y +7$ and $5y =12x + 6$. Find the angle between them and also the equations to the two striaght lines which pass through point $(4,5)$ and make equal angles with two lines. 


I can get the angles easily as $\pi - \tan^{-1}(\frac{33}{56})$ and $\tan^{-1}(\frac{33}{56})$, But the problem is the second part. 




  
    I know that if a point is given to me as $(x^{\prime} , y^\prime)$ and a line $y =mx + c$, then if we consider the line as the base of a isosceles triangle(it is the unequal side of that isosceles triangle) and the given point is the opposite to the line, then the slope of other two sides are $$ m^\prime = {\tan \alpha \pm m \over 1 \mp m\tan\alpha }$$ 
    where $\alpha$ is equal angle of the triangle. 
  


So to solve that question I supposed that the given two lines are the two equal sides of a isosceles triangle. With $\displaystyle\tan \alpha =  \tan\left({\pi - \tan^{-1}(\frac{33}{56}) \over 2}\right),  \tan\left({\tan^{-1}(\frac{33}{56}) \over 2}\right) = {11\over 3}, {3\over11}$

Since I know $m^\prime$ in this situation (I assume it to be ${12\over5}$ and ${3\over 4}$(Slopes of the given lines)).

so putting all this in the equation  

$$ \left({12\over5} \text{ or } {3\over 4}\right)= {\left({11\over 3} \text { or } {3\over11}\right) \pm m \over 1 \mp m\left({11\over 3} \text { or } {3\over11}\right) }$$

Solving for $m$ in the above equation gives $6$ different values of $m$ which directly implies $6$ different striaght line equations; Of which only two are correct.


How can I find these two out from these $6$ equations ?
Any easier way of doing this question ?




Here dotted lines are the desired equations and solid lines are the given equations.



(Click on the picture for the link to graphing calculator)
Let $y=m(x-4)+5$ be the equation of a generic line through $P=(4,5)$. Compute the intersection points $A$ and $B$ of that line with the two given lines (as a function of $m$). Find the squared distances of $A$ and $B$ from $O=(-59/33,-34/11)$, the common point of the given lines. Finally, set $AO^2=BO^2$ to find the values of $m$ which give an isosceles triangle.
I make you clear that I'm writing equation first as 

12y = 5x +6   ----(1)  instead of 5y =12x +6

And 2nd eqⁿ is 4y = 3x -7    ------(2) 

Here
m₁ = 5/12, m₂ =3/4. 

Let m is slop of desire line, then

y = m(x-4) +5   ------(3)

Now angle between (1)&amp;(3) will be

Θ₁ = |(12m -5)/(12 +5m)|

Similarly angle between (2)&amp;(3) will be

Θ₂ = |(4m -3)/(4 +3m)|

Now since Θ₁=θ₂ given,then

(12m -5)/(12 +5m) 
           = ± (4m -3)/(4 +3m)

After cross multiplication,we get 

48m+36m² -20-15m 

= ± (48m-36+20m²-15m)

On Taking +ve sign 

16m² = -16, which is unacceptable.

On taking -ve sign

96m+56m² -30m-56=0

⇒56m² +66m -56=0

⇒28m² +33m -28=0

⇒m = (-33±65)/56

    =32/56 or -98/56

    =4/7 or -7/4


Thus desired line would be

7y = 4x -11 &amp; 4y = -7x +33.