# Dealing with negative areas— coordinate geometry

Question:

Find the area of a quadrilateral in the Cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2)

My solution:

[I meant to draw B(0,7), instead, I drew B(7, 0), apologies. Though, the rest is still applicable.]

Area(ABCD) = Ar(ABC) + Ar(ADC)

Let A(-4, 5) be (x1, y1); B(0, 7) be (x2, y2); C(5, -5) be (x3, y3)

Applying the formula:

I got Ar(ABC) = -58/2

Now, let C and A have the same coordinates and D(-4, -2) be (x2, y2), I got Ar(ADC) = 63/2.

Now, Ar(ABCD) = Ar(ABC) + Ar(ADC)

= 58/2 + 63/2 (not -58/2 as area can't be negative)

= 121/2

My question: is my statement of area not having a negative value correct or have I done something wrong and that's why I'm getting a negative value? Because if I don't consider that area to be positive, I get a different, wrong, answer.

• You wrote $(0, 7)$, but you drew $(7, 0)$. – Sammy Black Nov 21 '14 at 4:44
• @SammyBlack Damn it! Lemme recalculate then... However, let's say hypothetically, had I gotten something like the figure above, would my reasoning be correct? – Always Learning Forever Nov 21 '14 at 4:47
• @SammyBlack Wait, I don't need to recalculate as the values I put in were correct, it's just the flaw in the diagram. – Always Learning Forever Nov 21 '14 at 4:51

Assuming the only issue you had was the negative sign and your figure has no reflex angles, yes, your answer should be correct - the area of the triangles is indeed the absolute value of the answer you get from your expression. It's a little hard to read what formulas you're using, but basically, if you check, the following will be true of what it outputs: $$\text{ar}(ABC)=-\text{ar}(ACB)$$ that is, if we go around the triangle in the opposite direction, we get a negative sign. That's just kind of how coordinate geometry has to go because most any expression you can think of that might reasonably talk of area (that doesn't invoke absolute values) will be negative for some point. We might say this gives an orientation to the plane, in that we are assigning (for instance) triangles with vertices in clockwise order to be positive and triangles with vertices in counterclockwise order to be negative (or vice versa).
Of course, it is possible that the area of a quadrilateral $ABCD$ is not equal to $$|\text{ar}(ABC)|+|\text{ar}(ADC)|$$ for instance when the angle $D$ is a reflex angle, this yields the wrong answer. However, the expression $$|\text{ar}(ABC)+\text{ar}(BCD)|$$ will always work because, in a sense, we preserve the orientation of our triangles with respect to each other (so, if they cancel, they were meant to - the overall answer might be of the wrong sign, but when we take an absolute value after addition, this works out).