Slope Intercept Form

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 \[\text{Slope of the line }(m)=\tan\theta\]

Slope Intercept Form

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.

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 \[y=mx\] \[\therefore y=\pm x\]


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