Chapter 7.3 — Exercise 7.3 — Area of Triangle

Splitting a Triangle Into Trapezoids

Given A(x₁,y₁), B(x₂,y₂), C(x₃,y₃), drop a perpendicular from each vertex down to the x-axis, landing at feet P, Q, and R. That construction breaks the triangle's area into three trapezoids, two added and one subtracted:

The absolute value bars at the front matter more than they look: depending on which order the three vertices are listed in, the expression inside can come out negative, but a physical area is never negative, so the final answer always takes the positive value regardless of vertex order. Labelling the vertices A, B, C in a different order around the triangle, or even starting from a different vertex altogether, only ever flips the sign of the bracket — it never changes the actual area once the absolute value is applied, which is exactly why the formula works no matter how the three points happen to be listed in a problem.

Area Zero Means No Triangle At All

This formula quietly hands over a second, entirely free fact: if three points are collinear, they can't enclose any area, so the area formula evaluates to exactly zero. Area = 0 is therefore a genuine collinearity test — a different route to the same conclusion Exercise 7.1 reached by comparing distances directly. Where the distance-based test needed three separate square-root calculations and then a comparison between them, the area-based test needs only one substitution and a check against zero — often the faster of the two, especially once a and b coefficients start involving fractions or decimals that make repeated square roots tedious.

Three Areas, Direct From the Formula

The third row is worth a direct sanity check: this triangle has a horizontal leg of length 3 along the x-axis and a vertical leg of length 2 along the y-axis, meeting at a right angle at the origin — so its area is also just ½×base×height = ½×3×2 = 3, matching the coordinate formula exactly and confirming it's really just Pythagoras-style right-triangle geometry underneath the algebra. The first row is worth a second glance too, purely as a bracket-arithmetic warning: three separate negative signs appear inside the working (−4−3, and the subtraction inside each pair of parentheses), and losing track of even one of them flips the final answer's sign before the absolute value has a chance to fix it — writing out each subtraction as its own small step, rather than combining several in one line, is the more reliable habit here.

Finding an Unknown That Forces Collinearity

Setting the area formula equal to zero, instead of computing a nonzero area, finds whatever unknown value makes three points collinear.

Every row here follows an identical two-step shape: substitute the three points into the area formula, then instead of simplifying to a number, set the whole bracket to zero and solve the resulting linear equation for K. The third row is the one worth double-checking by direct substitution afterward, since K appears in two different coordinates at once (the point (K,K) uses the same unknown for both x and y) — plugging K=7/3 back into all three original points and re-running the area formula is a genuinely fast way to confirm nothing was mismatched between the x-terms and y-terms during the algebra.

The Medial Triangle's Area Is Always a Quarter

Joining the midpoints of a triangle's three sides creates a second, smaller triangle inside it — the medial triangle. For the outer triangle (0,−1), (2,1), (0,3), the area works out to 4 square units. Its midpoints are D(1,2), E(0,1), and F(1,0), and the inner triangle DEF has area ½|1(1−0)+0(0−2)+1(2−1)| = 1 square unit.

This 1:4 ratio isn't a coincidence tied to these specific coordinates — it's a general property of every triangle, since each of the three smaller triangles formed by the midpoints is always congruent to the medial triangle, splitting the whole area into four equal pieces regardless of the original triangle's shape. It's worth remembering as its own fact rather than re-deriving it from scratch each time: given any triangle's area, the medial triangle's area is always exactly a quarter of it, without needing to find a single midpoint coordinate first.

A Quadrilateral, Split Along Its Diagonal

The area formula only works directly for triangles, but any quadrilateral splits cleanly into two triangles along either diagonal. For quadrilateral (−4,−2), (−3,−5), (3,−2), (2,3), taken in order, diagonal AC splits it into △ABC and △ACD:

Choosing the other diagonal, BD, would split the same quadrilateral into △ABD and △BCD instead — a different pair of triangles entirely, but one that adds up to the identical total area of 28 square units, since both diagonals cut through the same four-sided region without leaving anything out or double-counting anything either way.

Cross-Checking With Heron's Formula

A second, older formula finds a triangle's area purely from its three side lengths, without touching coordinates directly — useful as an independent check on the coordinate formula above.

For triangle (2,3), (6,3), (2,6): the three sides work out to 5, 3, and 4 — a familiar 3-4-5 right triangle. Here s=(5+3+4)/2=6, giving Area=√(6·1·3·2)=√36=6 square units. Checking the same triangle with the coordinate formula directly, ½|2(3−6)+6(6−3)+2(3−3)| = ½|−6+18+0| = 6, confirms both routes land on the identical answer — exactly the kind of two-method agreement worth trusting a result on, and a genuinely useful check whenever a problem happens to give both the coordinates and enough information to compute the side lengths as well.

From Area to Steepness

Every quantity measured so far in this chapter — distance, a dividing point's position, and now area — has come from comparing coordinates directly. Exercise 7.4 turns to a different kind of measurement entirely: not a length or a region, but the steepness of a line itself, found from the same pair of coordinates this whole chapter has been built on. For the collinearity idea this exercise's area-equals-zero test connects back to, revisit Exercise 7.1.