You have 2 possible line equations, for triangles sitting on top of the x-axis.
i'm assuming =n is short for equation.
Suppose the triangle sits on the positive half of the x-axis from zero to infinity.
If the perp from the origin to the line is 60 degrees.
then the line makes an angle of 30 degrees with the x-axis.
Be sure this is clear before proceeding,
draw a diagram if necessary.
\(\displaystyle tan30^o=\frac{1}{\sqrt{3}}\)
this gives us the ratio of height over base.
Therefore height=k and base=\(\displaystyle k\sqrt{3}\)
Be sure that you are clear about this before proceeding......\(\displaystyle \frac{k}{k\sqrt{3}}=\frac{1}{\sqrt{3}}\)
Now use the area to discover k.
Triangle area is \(\displaystyle 0.5(k)k\sqrt{3}=54\sqrt{3}\)
\(\displaystyle k^2=108\)
\(\displaystyle k=\sqrt{108}\)
The line slope \(\displaystyle m=-\frac{1}{\sqrt{3}}\)
\(\displaystyle c=k=\sqrt{108}\)
The equation is y=mx+c.
What about working out the 2nd one?
