Question

Find the equation of the lines for which $tan\,\theta=\frac{1}{2}$, where $\theta$ is the inclination of the line and $\left(i\right) y$-intercept is $-\frac{3}{2} \left(ii\right) x$-intercept is $4$.

	$(i)\quad$	$(ii)$
(a)$\quad$	$y - 2x + 3 = 0$$\quad$	$2y - x + 4 = 0$$\quad$
(b)$\quad$	$2y - x + 3 = 0\quad$	$2y - x + 4 = 0\quad$
(c)$\quad$	$2y - x + 3 = 0\quad$	$y - 2x + 4 = 0\quad$
(d)$\quad$	$y - 2x + 3 = 0$	$y - 2x + 4 = 0$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (B)

$(i)$ Here, slope of the line is $m=tan\,\theta=\frac{1}{2}$ and
$y$-intercept $\left(c\right) =-\frac{3}{2}$. Therefore, by slope-intercept form, the equation of the line is
$y=\frac{1}{2}x-\frac{3}{2}$ or $2y-x+3=0$
$\left(ii\right)$ Here slope of line is $m=\frac{1}{2}$ and $x$-intercept $\left(d\right) = 4$
By slope-intercept form, the equation of the line is
$y=\frac{1}{2}\left(x-4\right)$ or $2y-x+4=0$.