Area Triangle: Find Formula in Terms of $p, q, r$

[anemone]

Find in terms of $p,\,q$ and $r$, a formula for the area of a triangle whose vertices are the roots of $x^3-px^2+qx-r=0$ in the complex plane.

 

[jvicens]

The area of a triangle can be calculated using the formula $A = \frac{1}{2}bh$, where $b$ is the base of the triangle and $h$ is the height. In this case, the base of the triangle would be the distance between two of the roots and the height would be the distance from the third root to the line connecting the other two roots.

Let us label the roots as $a, b,$ and $c$. The base of the triangle would be $|a-b|$ and the height would be $|c-\frac{a+b}{2}|$. Therefore, the area of the triangle would be:

$A = \frac{1}{2}|a-b||c-\frac{a+b}{2}|$

Using the quadratic formula, we can express the roots $a$ and $b$ in terms of $p$, $q$, and $r$:

$a = \frac{p+\sqrt{p^2-4q}}{2}$

$b = \frac{p-\sqrt{p^2-4q}}{2}$

Substituting these values into the formula for the area, we get:

$A = \frac{1}{2}\left|\frac{p+\sqrt{p^2-4q}}{2}-\frac{p-\sqrt{p^2-4q}}{2}\right|\left|c-\frac{\frac{p+\sqrt{p^2-4q}}{2}+\frac{p-\sqrt{p^2-4q}}{2}}{2}\right|$

Simplifying this, we get:

$A = \frac{1}{2}\left|\sqrt{p^2-4q}\right|\left|c-\frac{p}{2}\right|$

Now, we need to express the third root $c$ in terms of $p$, $q$, and $r$. From the cubic equation, we can see that:

$c = \frac{r}{a\cdot b}$

Substituting the values for $a$ and $b$ from earlier, we get:

$c = \frac{r}{\left(\frac{p+\sqrt{p^2-4q}}{2}\right)\left(\frac{p-\sqrt{p^2-4q}}{2}\right)}$

Simplifying this,