Question

Write the equation of the lines for which $\tan \theta =\frac{1}{3},\text{where}\theta$ is the inclination of a line with positive x-axis and also making x-intercept equal to 4 ?

• A. x + 3y + 12 = 0
• B. x - 3y - 2 = 0
• C. x - 3y + 12 = 0
• D. x - 3y + 2 = 0

Answer 1

The correct option is C x - 3y + 12 = 0

The equation of line with slope, $m=\tan \theta =\frac{1}{3}$ and y-intercept is 4 is

$y=mx+c$

$\Rightarrow y=\frac{1}{3}\times x+4\Rightarrow y=\frac{x+12}{3}$

$\Rightarrow 3y=x+12$

$\therefore x-3y+12=0$

Hence, option c is correct.