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 $4$.

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

• 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 $2y-x+4=0$.