Question

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

• A. $9x-7y=1;7x+9y=73$
• B. $2x-2y=1;5x+7y=45$
• C. $3x-6y=4;3x+4y=14$
• D. $4x-2y=1;6x-8y=63$

Answer 1

The correct option is A $9x-7y=1;7x+9y=73$

Let the Slope of line be $m$

Now, slope of other lines are $\frac{3}{4},\frac{12}{5}$

The Slope is Equally inclined so it must lie between these slopes

such that

$\frac{3}{4}<m<\frac{12}{5}$

As it is equally inclined so angle between the line and the two given line must be same

$\Rightarrow {\theta }_1={\theta }_2\Rightarrow \tan {\theta }_1=\tan {\theta }_2\Rightarrow \frac{\frac{12}{5}-m}{1+\frac{12}{5}m}=\frac{m-\frac{3}{4}}{1+\frac{3}{4}m}\Rightarrow 63m^2-32m-63=0$

$\Rightarrow m=\frac{32\pm \sqrt{{32}^2+4\ast 63\ast 63}}{2\ast 63}\Rightarrow m=\frac{-7}{9},\frac{9}{7}$

Now, Equation of line as it passes through $(4,5)$ will be

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

Putting the obtained values of $m$ in above Equation we get

$9x-7y=17x+9y=73$