Analytic analyse of triangle

Question

If the area of triangle ABC with the vertices $A(3,1)$, $B(-1,-2)$ and $C(m,1)$ is five square units find the possible values of m?
I found the answer to be 19/3 but actual answer is 19/3 and -1/3 I solved the question using area of a triangle formula why is there -1/3 ?

Answer 1

$C$ can be on either side of line $AB$. Consider a modulus (absolute value) of the triangle's area.

Added

The area of a triangle is half the length of a vector product of two vectors defined as the triangle's sides. So, for example, $$S_{\triangle ABC} = \frac 12\left|\vec{AB}\times\vec{AC}\right|$$ In this case we know area is $5$ units: $$S_{\triangle ABC} = 5.$$ As we (hopefully) know, $$\left|\vec{AB}\times\vec{AC}\right|=|x_{AB}y_{AC}-y_{AB}x_{AC}|= |(x_B-x_A)(y_C-y_A)-(y_B-y_A)(x_C-x_A)|$$ $$=|((-1)-3)(1-1)-((-2)-1)(m-3)|=|(-4)\cdot 0 + 3(m-3)| = 3|m-3|$$ Plugging that to the above requirement we get $$\frac 12\cdot 3|m-3|=5$$ that is $$|m-3|=\frac {10}3$$ hence either $m-3=\frac {10}3$ or $m-3=-\frac {10}3$.

As a result either $m=3+\frac {10}3=\frac{19}3$ or $m=3-\frac {10}3=-\frac 13$.