# Equation of a straight line passing through passing through a point and equally inclined to two other lines

Find the equation of the straight line passing through the point (4,5) and equally inclined to the lines $3x= 4y+7$ and $5y=12x+6$.

I know that the equation of the bisector is given by: $\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}$=$\pm$$\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}} but I am facing real difficulty in finding which sign I should choose and why? the answers are: 9x-7y=1 and 7x+9y=73 • I don't understand: "equally inclined to the lines". – Guillemus Callelus Sep 28 '17 at 16:08 • @GuillemusCallelus the angle bisector of two lines is equally inclined to the two lines – Abcd Sep 28 '17 at 16:09 • Choose the one which passes through (4,5) – Math Lover Sep 28 '17 at 16:10 • This is basically a duplicate of your previous question, albeit with specific lines and points. If you didn’t really understand the answer that you got, why did you accept it? – amd Sep 28 '17 at 19:36 ## 3 Answers By using your idea we can get two slops: m=\frac{9}{7} or m=-\frac{7}{9} and from here two equations:$$y-5=\frac{9}{7}(x-4)y-5=-\frac{7}{9}(x-4)$$• Which "idea"of mine are you referring two? – Abcd Sep 28 '17 at 16:24 • @Abcd I mean the following your idea: \dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm$$\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$ – Michael Rozenberg Sep 28 '17 at 16:25
• Please verify : $\dfrac{3x-4y-7}{5}= \pm \dfrac{12x-5y+6}{13}$ – Abcd Sep 28 '17 at 16:31
• @Abcd It's exactly! – Michael Rozenberg Sep 28 '17 at 16:33
• But I got slope = $-7/9$ using this. – Abcd Sep 28 '17 at 16:33

the straight line which passes through $$P(4;5)$$ has the equation $$y=m(x-4)+5$$ converting the others into the Hessian Normalform we get $$0=\frac{4y-3x+7}{\pm5}$$ and the other $$0=\frac{5y-12x-6}{\pm 13}$$ the $$Point (0; -4m+5)$$ is situated on our line and we must compute $$\frac{|5(-4m+5)-6|}{13}=\frac{|4(-3m+5)+7|}{5}$$ fromk here you will get $m$

You have TWO lines which form equal angles with the given straight lines

$3x-4y-7=0;\;12x-5y+6=0$

$\dfrac{3x-4y-7}{\sqrt{3^2+4^2}}=\pm\dfrac{12x-5y+6}{\sqrt{12^2+5^2}}$

$\dfrac{3x-4y-7}{5}=\pm\dfrac{12x-5y+6}{13}$

$13(3x-4y-7)=\pm 5(12x-5y+6)$

$99 x-77 y-61=0;\;21 x+27 y+121=0$

$y=\dfrac{9 x}{7}-\dfrac{61}{77};\;y=-\dfrac{7 x}{9}-\dfrac{121}{27}$

If the wanted lines gave to pass through the point $(4,5)$ then their equation is

$y-5=m(x-4)$

where $m_1=\dfrac{9}{7};\;m_2=-\dfrac{7}{9}$

Hope this helps