How to find third point of triangle with specific area?

Question

If $A = (3, 5)$ and $B = (11, 11)$ are fixed points, find the point(s) $P$ on the $x$-axis such that the area of the triangle $ABP$ equals $30$.

I feel that somehow idea of cross product is involved, but I don't know how.

$\frac{A\times B}{2}$ is area of triangle between vectors $A$ and $B$.

Answer 1

There are a couple of ways you can use cross product...

you can say $A = \frac 12 |(3,11,x)\times (5,11,0)| = 30$

or you can say:

$A = \frac 12 |(3-x, 5)\times(11-x,11)| = 30$

The shoelace algorithm is essentially the first one.

$\begin{array} {} 3&5\\11&11\\x&0\end{array}$

$A = \frac 12 |(3\cdot 11 +11\cdot 0 + x\cdot 5 - 5\cdot 11 - 11\cdot x - 0\cdot 3)| =30\\ |22 + 6x| = 60\\ x = \frac {38}{6}\\ x = -\frac {82}{6}$

Answer 2

Hint:

The area of a triangle with vertices $(x_A,y_A)$, $(x_B,y_B)$ and $(x_C,y_C)$ is given by $$\frac12\left|\det\begin{pmatrix}x_A&x_B&x_C\\y_A&y_B&y_C\\1&1&1\end{pmatrix}\right|$$

Then, if $(x_C,y_C)$ lies oh the $x$ axis, $y_C=0$ and then we are looking for those $x_C$ such that $$\frac12\left|\det\begin{pmatrix}3&11&x_C\\5&11&0\\1&1&1\end{pmatrix}\right|=30$$