Write the equation of the lines for which $\tan \theta = \frac{1}{2}$,where $\theta$ is the inclination of the line and $y$-intercept is $-\frac{3}{2}$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(A) The slope of the line is given by $m = \tan \theta = \frac{1}{2}$.
The $y$-intercept is given by $c = -\frac{3}{2}$.
Using the slope-intercept form of a line,$y = mx + c$,we substitute the values:
$y = \frac{1}{2}x - \frac{3}{2}$.
Multiplying the entire equation by $2$,we get:
$2y = x - 3$.
Rearranging the terms,we get the equation of the line as:
$x - 2y - 3 = 0$.

Explore More

Similar Questions

The equation of the straight line whose slope is $\frac{-2}{3}$ and which divides the line segment joining $(1, 2)$ and $(-3, 5)$ in the ratio $4:3$ externally is

The equation of a given straight line is $\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma$. If the equation of the line perpendicular to the given line and passing through $(\alpha, \beta)$ is $\frac{x}{a}+\frac{y}{b}=1$,then $\frac{b}{a}$ is equal to

Find the $y$-intercept of a line that is perpendicular to $3x + y = 3$ and passes through the point $(2, 2)$.

Find the slope of the line,which makes an angle of $30^{\circ}$ with the positive direction of the $y$-axis measured anticlockwise.

$A$ straight line passing through the point $(2, 2)$ intersects the lines $\sqrt{3}x + y = 0$ and $\sqrt{3}x - y = 0$ at points $A$ and $B$ respectively. Find the equation of the line $AB$ such that the triangle $OAB$ is an equilateral triangle,where $O$ is the origin.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo