I do not know where my line intercepts the $y-$axis. I know it intercepts the x axis at coordinate $(10, 0)$. At an angle of $30^{\circ}$.
So would $$y = mx + b = (10\tan(30)) + b$$
be correct? How do I find where my line intersects the $y-$axis and find the equation for this line?
On
Hint:
If you know the slope $m$ and any point $(x_0,y_0)$ on the line, then by solving
$$y_0=mx_0+p,$$ you get $p$.
On
Your equation $$y = mx + b = (10\tan(30)) + b$$ is not correct.
Note that the slope of the line is $\tan(30)$ and you have the point $(10,0)$ on the line.
Thus the equation of the line is $$y-0=(\tan(30))(x-10)$$ which simplifies to $$ y = (\frac {\sqrt 3}{3})(x-10)$$
On
Clearly you’re familiar with the slope-intercept form of the equation $y=mx+b$. For $m=0$, there’s an equivalent form that uses the reciprocal of the slope and $x$-intercept $c$: $x=\frac1m y+c$. You should convince yourself that this is correct by converting it to the more familiar slope-intercept form.
$m=\tan(30^\circ)$Then you know that $(10,0)$ is on the line, so $10 \tan(30^\circ)+b=0$, or $b=-10\tan(30^\circ)$. Therefore the equation of the line is $$y=\tan(30^\circ)(x-10)$$