# Slope Intercept Form

Here, we will discuss about the slope intercept form of the equation of straight lines.

Let a straight line intersect x-axis and y-axis at $A$ and $B$ respectively. Let $AB$ make an angle $\theta$ with the positive x-axis. Then $\text{ Slope of the line }(m)=\tan\theta$

Let $OB=c$ and $P(x,y)$ be any point on the line. Draw $PM$ perpendicular to x-axis and $BN$ parallel to x-axis. Then, $\tan\theta=\frac{NP}{BN}$ $m=\frac{PM-MN}{OM}$ $m=\frac{y-c}{x}$ $mx=y-c$ $\therefore y=mx+c$

Since this equation is satisfied by the coordinates $(x,y)$ of any point $P$ on the line, it is the equation of the line. This equation is referred to as the slope intercept form or simply the slope form of an equation of a straight line.

• If the line passes through the origin, $c=0$, and the equation will be $y=mx$.
• If the line be parallel to the x-axis (slope $m=0$), the equation becomes $y=c$.
• If the line be parallel to the y-axis, then the equation cannot be written in this form.

### Find the equation of a line making an angle $60°$ with the positive x-axis and cutting an intercept $3$ from the y-axis.

Here, $\text{y-intercept } (c)=3$ $\theta=60°$ $\therefore \text{Slope }(m)=\tan 60°=\sqrt{3}$

Thus, equation of straight line is $y=mx+c$ $\therefore y=\sqrt{3}\text{ }x+3$

### Find the equation of a straight line making an angle of $135°$ with the positive x-axis and cutting an intercept $3$ from the negative y-axis.

Here, $\text{y-intercept }(c)=-3$ $\theta=135°$ $\therefore\text{Slope }(m)=\tan 135°=-1$

Thus, equation of straight line is $y=mx+c$ $y=-x-3$ $\therefore y+x+3=0$

### Find the equation of the straight lines bisecting the angle between the axes.

Since the line bisects the angle between the axis, $\theta=45°\text{ or }135°$. $\therefore\text{Slope }(m)=\tan 45°\text{ or }\tan 135°$ $=1\text{ or }-1$ $=\pm 1$

Thus, equation of the line is $y=mx$ $\therefore y=\pm x$