Question

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

• A. $9x-7y=1$
• B. $9x+7y=71$
• C. $7x+9y=73$
• D. $7x-9y+17=0$

Answer 1

The correct option is A $9x-7y=1$

Multiple options are correct.

$3x=4y+7$

$\to y=\frac{3}{4}x-\frac{7}{4}\Rightarrow m_1=\frac{3}{4}$

$\therefore \tan x=\frac{m-\frac{3}{4}}{1+\frac{3}{4}m}=\frac{4m-3}{4+3m}$........(1)

$5y=12x+6$

$\to y=\frac{12}{5}x+\frac{6}{5}\Rightarrow m_2=\frac{12}{5}$

$\therefore \tan x=\frac{m-\frac{12}{5}}{1+\frac{12}{5}m}=\frac{5m-12}{5+12m}$........(2)

comparing (1) and (2), we get

$\frac{4m-3}{4+3m}=\frac{5m-12}{5+12m}$

solving the above equations, we get,

$m=\frac{9}{7},\frac{-7}{9}$

$(y-y_1)=m(x-x_1)$

given, $(4,5)$, we have

$y-5=\frac{-7}{9}(x-4)$

$\Rightarrow 9y+7x=73$.........A

$y-5=\frac{9}{7}(x-4)$

$\Rightarrow 9x-7y=1$..........B