Question

Find the equation of the line which pass through $(4,5)$ and makes equal angles with the lines $5y=12x+6$ and $3x=4y+7$.

Answer 1

Let the equation of required line be
$y=mx+c$... (1)
Given line is $5y=12x+6$
$\Rightarrow y=\frac{12}{5}x+\frac{6}{5}$ .. ....(2)
& $3x=4y+7$
$\Rightarrow y=\frac{3}{4}x-7$......(3)
Slope of (2) $=\frac{12}{5}=m_1$
Slope of (3) $=\frac{3}{4}=m_2$

Let ${\theta }_1=$ Angle between (1) & (2)
$\tan {\theta }_1=\frac{m-\frac{12}{5}}{1+\frac{12m}{5}}=\frac{5m-12}{5+12m}$
Let ${\theta }_1=$ Angle between (1) & (3)
$\tan {\theta }_2=\frac{m-\frac{3}{4}}{1+\frac{3m}{4}}=\frac{4m-3}{4+3m}$

Since${\theta }_1={\theta }_2$
$\frac{5m-12}{5+12m}=\frac{4m-3}{4+3m}$
As same line is inclined equally to other two lines $\tan {\theta }_1=-\tan {\theta }_2$

$20m+15m^2-48-36m=-(20m-48m^2+15+36m)$
$15m^2+48m^2-16m-16m-48-15=0$
$63m^2-32m-63=0$
$m=\frac{32\pm \sqrt{(-32{)}^2-4\times 63\times -63}}{2\times 63}=\frac{32\pm 130}{126}$
$=\frac{162}{126}or-\frac{98}{126}$
$=\frac{9}{7}or-\frac{7}{9}$

$\therefore$ Equation line is $y=\frac{9}{7}x+c$ or $y=-\frac{7}{9}x+c$
However since slopes of lines (2) & (3) are positive, slope of the required line also positive.
$\therefore y=\frac{9}{7}x+c$
Now since (4, 5) lies on this line
$5=\frac{9}{7}\times 4+c$
$\Rightarrow c=5-\frac{36}{7}=-\frac{1}{7}$
$\therefore$ Equation of line is $y=\frac{9}{7}x-\frac{1}{7}$
$\Rightarrow y=\frac{9x-1}{7}$
$\Rightarrow 7y=9x-1$
$\Rightarrow 9x-7y=1$
Hence, the equation of required line is $9x-7y=1$.