Question

The equation of the straight line cutting off an intercept $8$ on x-axis and making an angle of ${60}^o$ with the positive direction of y-axis is

• A. $x-\sqrt{3}y-8=0$
• B. $x+\sqrt{3}y=8$
• C. $y-\sqrt{3}y=8$
• D. $y+\sqrt{3}y=8$

Answer 1

The correct option is A $x-\sqrt{3}y-8=0$

Let the equation of line be

$y=mx+c$

Now at $y=0,x=8$

$\therefore 0=8m+c$

$\therefore C=-8m$

$\therefore \frac{dx}{dy}=\tan 60$

$\therefore \frac{dy}{dx}=\frac{1}{\sqrt{3}}$

$\therefore m=\frac{1}{\sqrt{3}}$

$\therefore C=\frac{-8}{\sqrt{3}}$

$\therefore y=\frac{x}{\sqrt{3}}-\frac{-8}{\sqrt{3}}$

$\therefore \sqrt{3}y=x-8$

$\therefore x-\sqrt{3}y-8=0$