Visual walkthrough — Area of triangle using coordinate formula

Before any symbol appears, let us fix the vocabulary.

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Step 1 — Put the three pins down and drop them to the floor

WHAT. Place $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$. From each one, draw a vertical line straight down to the x-axis.

WHY. Because a shape sitting flat on the floor is the easiest shape whose area we know how to compute. We are going to trade the awkward slanted triangle for a few tidy floor-standing trapezoids.

PICTURE. Look at the three dotted vertical drops. Each drop hits the floor at $(x_i,0)$ — same $x$, but $y=0$. The three drops chop the region under the triangle into vertical strips.

In the picture, $x_1$ is how far right the drop of $A$ lands, and $y_1$ is how tall $A$ stands above the floor. Nothing here is a formula yet — just names for distances.

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Step 2 — One edge makes one trapezoid; find its area

WHAT. Take the edge from $A$ to $B$. Its two endpoints have heights $y_1$ and $y_2$. Its two drops land at $x_1$ and $x_2$. Together with the floor, they bound a trapezoid.

WHY use a trapezoid area, and why this one? A trapezoid with two vertical parallel sides has the friendliest area rule of any four-sided shape: average of the two heights, times the horizontal width. We choose it because our two vertical drops are exactly those two parallel sides — the shape was hand-picked to fit this rule.

PICTURE. The left wall has height $y_1$, the right wall has height $y_2$, and the floor gap between them has width $x_2-x_1$.

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Step 3 — Why the width carries a sign (this is the secret)

WHAT. Keep the same formula $\frac{1}{2}(y_a+y_b)(x_b-x_a)$, but now stare at the factor $(x_b-x_a)$ under different edges.

WHY. In the naive picture you might worry "what if $B$ is to the left of $A$?" The genius is that we do not need to worry. The subtraction $x_b-x_a$ handles left-vs-right by itself: it goes negative automatically. That negative sign will do the subtracting for us, so we never have to decide which trapezoids to add and which to remove.

PICTURE. Two copies of the same edge: one walked rightward (width positive, strip counted +), one walked leftward (width negative, strip counted −). Same strip, opposite sign, purely because of walking direction.

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Step 4 — Walk the whole loop $A\to B\to C\to A$

WHAT. Write down all three edge-trapezoids, each with its own built-in sign, and add them.

WHY. We are closing the loop. Some edges live above the triangle and some below it. When you add rightward-positive and leftward-negative strips all the way around a closed loop, every square metre outside the triangle gets added once and subtracted once — it cancels. Only the enclosed triangle survives.

PICTURE. The three strips shaded; arrows show the walking direction on each edge. The overlapping outside regions are marked "cancels".

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Step 5 — Multiply out and watch the junk cancel

WHAT. Expand each of the three products and collect terms.

WHY. Right now the formula is correct but ugly. Expanding reveals that the "self" terms — the ones pairing $x_i$ with its own $y_i$ — appear twice with opposite signs and vanish. What is left is short and memorable.

PICTURE. A term-by-term board: the three diagonal pairs $x_1{y}_1$, $x_2{y}_2$, $x_3{y}_3$ each show up as $+$ and $-$, struck through.

Expanding each bracket (drop the common $\frac{1}{2}$ for a moment): $(y_1+y_2)(x_2-x_1)=x_2{y}_1+\cancel{x_2{y}_2}-\cancel{x_1{y}_1}-x_1{y}_2$ $(y_2+y_3)(x_3-x_2)=x_3{y}_2+\cancel{x_3{y}_3}-\cancel{x_2{y}_2}-x_2{y}_3$ $(y_3+y_1)(x_1-x_3)=x_1{y}_3+\cancel{x_1{y}_1}-\cancel{x_3{y}_3}-x_3{y}_1$

• Each $\cancel{x_i{y}_i}$ appears once $+$ and once $-$ → gone.
• What survives, grouped by each $x_i$: $x_1(y_3-y_2)+x_2(y_1-y_3)+x_3(y_2-y_1)$

This is exactly the negative of the standard expression $x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$. A sign we are about to throw away anyway.

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Step 6 — Take absolute value: orientation is not area

WHAT. Put the $\frac{1}{2}$ back and wrap the whole thing in $|\cdots |$.

WHY. The signed sum $S$ is positive if we walked counter-clockwise, negative if clockwise. Area is a how-much-paper quantity — it cannot be negative. The bars $|\cdots |$ discard the walking direction and keep only the size.

PICTURE. Same triangle traversed two ways: CCW gives $+S$ (green), CW gives $-S$ (red). Both give the same $|S|$.

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Step 7 — The degenerate case: when the triangle is flat

WHAT. Feed in three points on one straight line: $A(1,2)$, $B(2,4)$, $C(3,6)$.

WHY. A good formula must not just work — it must tell you when there is no triangle. If the three pins line up, the "triangle" is a squashed sliver of zero width.

PICTURE. The three strips exactly overlap; there is no leftover region. Signed sum $=0$.

$\text{Area}=\frac{1}{2}|1(4-6)+2(6-2)+3(2-4)|=\frac{1}{2}|-2+8-6|=\frac{1}{2}|0|=0$

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Step 8 — Sign check with real numbers

WHAT. Run one ordinary triangle and one negative-coordinate triangle, watching the sign inside the bars.

WHY. To see that the inside can be negative (clockwise ordering) yet the answer is still a clean positive area.

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The one-picture summary

This last figure compresses everything: three drops build three signed floor-strips; walk the loop; outside cancels; halve; take the size.

Recall Feynman retelling — say it to a friend

Q: You know only the three corners. How do you get the area? Corners to area with only coordinates ::: Walk around the triangle corner to corner. Under each side, look at the strip going straight down to the floor. Walking rightward, count the strip plus; walking leftward, count it minus. Add all three strips: the outside bits cancel and you are left with the triangle.

Q: Why is each strip's area $\frac{1}{2}(y_a+y_b)$ times the width? Strip area rule ::: It is a trapezoid standing on the floor with vertical walls of height $y_a$ and $y_b$; a trapezoid's area is the average wall height times the width.

Q: Why do we never worry about which corner is leftmost? Why order does not matter ::: The width $x_b-x_a$ goes negative all by itself when you walk leftward, so the subtractions happen automatically.

Q: Why the $|\cdots |$ at the end? Why absolute value ::: The bare sum is positive for a counter-clockwise walk and negative for clockwise. Area cannot be negative, so we keep only the size.

Q: What does a zero answer mean? Meaning of zero ::: The three points are on one straight line — the triangle is flat, no area, they are collinear.

Recall Quick self-test

Area of $O(0,0),P(5,0),Q(0,12)$ ::: $\frac{1}{2}|0+60+0|=30$ — matches $\frac{1}{2}\cdot 5\cdot 12$. The factor out front is always ::: $\frac{1}{2}$, from the trapezoid average-height rule.