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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 $7x+9y=73$

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  • $\begingroup$ I don't understand: "equally inclined to the lines". $\endgroup$ – Guillemus Callelus Sep 28 '17 at 16:08
  • $\begingroup$ @GuillemusCallelus the angle bisector of two lines is equally inclined to the two lines $\endgroup$ – Abcd Sep 28 '17 at 16:09
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    $\begingroup$ Choose the one which passes through $(4,5)$ $\endgroup$ – Math Lover Sep 28 '17 at 16:10
  • $\begingroup$ 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? $\endgroup$ – amd Sep 28 '17 at 19:36
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By using your idea we can get two slops: $m=\frac{9}{7}$ or $m=-\frac{7}{9}$ and from here two equations: $$y-5=\frac{9}{7}(x-4)$$ $$y-5=-\frac{7}{9}(x-4)$$

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  • $\begingroup$ Which "idea"of mine are you referring two? $\endgroup$ – Abcd Sep 28 '17 at 16:24
  • $\begingroup$ @Abcd I mean the following your idea: $\dfrac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm$$\dfrac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$ $\endgroup$ – Michael Rozenberg Sep 28 '17 at 16:25
  • $\begingroup$ Please verify : $\dfrac{3x-4y-7}{5}= \pm \dfrac{12x-5y+6}{13}$ $\endgroup$ – Abcd Sep 28 '17 at 16:31
  • $\begingroup$ @Abcd It's exactly! $\endgroup$ – Michael Rozenberg Sep 28 '17 at 16:33
  • $\begingroup$ But I got slope = $-7/9$ using this. $\endgroup$ – Abcd Sep 28 '17 at 16:33
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the straight line which passes through $$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 $m$

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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

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