# Coordinate geometry prove for equilateral triangle area

Equilateral triangle OAB is drawn such that vertex O is in the origin and vertex B passes though line $ax+by+c=0$. So that $OAB\angle =OBA\angle = 60^\circ$. Prove the area of triangle is $\frac {c^2} {(\sqrt3) (a^2+b^2)}$ I tried to solve this problem by taking $B\equiv(\alpha,\beta)$, then gradient of $OB=\frac \beta\alpha$ then I determined gradient of $BA$ (say $m$) using the following theorem: $$\tan 60^\circ=\left|\frac {m-\frac\alpha\beta}{1+\frac{m\beta}{\alpha}}\right|$$ From this gradient I was able to express line equation of $AB$. Then using the perpendicular bisector of $OB$ , I tried to find the coordinates of $A$. But this gives a very large expression for coordinates of A.

When you shift $B$ along the line, the distance $|OB|$ changes; and if the triangle's side changes, so changes the area (with a square of the side length).
Hence the area can not depend on $a,b,c$ only.

However, you might be interested in a minimum possible area of a triangle with the line given. In such case find the minimum possible length $|OB|$ for $B$ on the line and calculate the triangle's area for that side length.

• Well explained. I got this question from a school test paper. There must have been a error in the question. Its the minimum area they are asking. Appreciated for sorting it out! – emil Apr 5 '18 at 16:14
• The minimum area will have a coefficient of $\frac{\sqrt 3}{4}$ instead of $\frac {1 }{ \sqrt 3}$ – WW1 Apr 5 '18 at 17:56
• @WW1 You're right. So there are two errors in the problem statement. – CiaPan Apr 5 '18 at 19:13

Do you have to edit, dear friend, because your problem has been deficiently stated. For example nothing prohibit that the line is $x = 0$ in which case $a=1,\space \space b=c=0$ so the given formula gives area $0$.

On the other hand, for all point $A$ in the line $y=\dfrac{\sqrt3\space x}{3}$, one has an equilateral triangle with side $\overline{OA}$ and base $\overline{OB}$ in the y-axis.

• Both lines go through $O$ so both have $c=0$ hence the given formula gives $0$ in both cases. And this is actually yhe minimu area in both cases. Anyway, for both you can find a triangle with arbitrarily large area – CiaPan Apr 5 '18 at 19:10
• The statement don't speak about a minimum. That's why I wrote "Do you have to edit". Regards. – Piquito Apr 6 '18 at 0:02
• But the given formula gives $0$ in both cases, so there's no 'other hand' here. – CiaPan Apr 6 '18 at 5:48