WHY does this work? Let's build it from scratch using the trapezoid method.
Step 1: Enclose the triangle using trapezoids to the x-axis
Drop vertical lines from each vertex to the x-axis. Each edge of the triangle, together with the two vertical lines and the x-axis segment beneath it, forms a trapezoid. The signed sum of these trapezoids equals the triangle's area.
Step 2: Set up the signed trapezoid sum (going around A→B→C→A)
We traverse the triangle's boundary in order. For each directed edge from (xa,ya) to (xb,yb), the trapezoid contributes a signed area:
21(ya+yb)(xb−xa)
WHY signed, and WHY this handles any ordering? The factor (xb−xa) is automatically negative when we move leftward and positive when we move rightward. So we do not need to assume x1≤x2≤x3. As we walk the closed loop A→B→C→A, the rightward edges add area and the leftward edges subtract it, and the leftover is exactly the enclosed triangle — for any vertex arrangement. If we walk counterclockwise the total is positive; clockwise gives negative. That is why we take ∣⋅∣ at the end.
Step 3: Write the three directed-edge contributions
Step 4: Sum all three (each already carries its own sign)
Signed Area=21[(y1+y2)(x2−x1)+(y2+y3)(x3−x2)+(y3+y1)(x1−x3)]
Note: all three terms are ADDED — the signs are already built into the (xb−xa) factors. (This is the crucial fix: it is a sum around the loop, not "add two, subtract one.")
Step 5: Multiply out and simplify
Expand each product:
(y1+y2)(x2−x1)=x2y1+x2y2−x1y1−x1y2(y2+y3)(x3−x2)=x3y2+x3y3−x2y2−x2y3(y3+y1)(x1−x3)=x1y3+x1y1−x3y3−x3y1
Add them. The ±x1y1, ±x2y2, ±x3y3 diagonal terms cancel in pairs. What survives:
x2y1−x1y2+x3y2−x2y3+x1y3−x3y1
Now group by each xi:
=x1(y3−y2)+x2(y1−y3)+x3(y2−y1)
This is the negative of the target expression, i.e. it equals −[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]. Since we take absolute value, the sign is irrelevant:
Area=21x1(y2−y3)+x2(y3−y1)+x3(y1−y2)✓
Take absolute value to ensure positive area regardless of vertex order. ✓
Recall Feynman Technique: Explain to a 12-year-old
Imagine you have three pins on a board at different spots. How much paper would you need to cover the triangle they make?
If you only know where the corners are (like "3 steps right, 2 steps up"), can you figure out the area without actually drawing it?
Yes! Here's the trick: walk around the triangle from corner to corner. For each side, look at the strip below it down to the floor (x-axis). If you walk rightward, count that strip as positive; if you walk leftward, count it as negative. When you finish the loop and add up all the strips, the extra bits cancel and you're left with exactly the triangle. Because rightward/leftward automatically flips the sign, it works no matter where the corners are!
The weird part with x₁(y₂ - y₃) is really saying "look at corner 1's x-position, and see how much the y-positions of the other two corners differ." Combine all three corners this way, divide by 2, and put bars around it (so the answer is positive) — done!
Heron's Formula - alternative area method using side lengths
#flashcards/maths
What is the coordinate formula for the area of a triangle with vertices A(x₁,y₁), B(x₂,y₂), C(x₃,y₃)?
Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
Why must we use absolute value in the coordinate area formula?
Because the expression can be negative (depending on vertex order—clockwise vs counterclockwise), but area is always a positive magnitude.
In the trapezoid derivation, why don't we need to assume x₁≤x₂≤x₃?
Because each directed edge contributes ½(yₐ+y_b)(x_b−xₐ), and (x_b−xₐ) is automatically negative for leftward travel and positive for rightward travel — the signs self-correct for any ordering as we loop around A→B→C→A.
What does it mean if the coordinate area formula gives exactly 0?
The three points are collinear (lie on the same straight line), so no triangle exists.
In the formula x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂), what is the pattern for subscripts?
Each x multiplies the difference of the OTHER TWO y-coordinates, cycling through (1→2→3→1): x₁ uses y₂,y₃; x₂ uses y₃,y₁; x₃ uses y₁,y₂.
What is the determinant form of the triangle area formula, and what does the determinant equal geometrically?
Area = ½|det(x₁,y₁,1,[x₂,y₂,1],[x₃,y₃,1]])|; the determinant equals twice the SIGNED AREA of the triangle (not a 3D volume) — the "1" column is a bookkeeping device.
If a triangle has vertices O(0,0), P(a,0), Q(0,b), what is its area using the coordinate formula?
Area = ½|0(0-b) + a(b-0) + 0(0-0)| = ½|ab| = ½ab, which matches the right-triangle formula ½×base×height.
What common mistake involves the ½ factor?
Forgetting to include ½ in front of the absolute value, leading to double the correct area.
How does the coordinate formula "detect" colinearity?
When three points are collinear, the signed trapezoid contributions around the loop perfectly cancel, yielding 0.
Dekho, jab tumhe teen points ke coordinates pata hain - A(x₁,y₁), B(x₂,y₂), aur C(x₃,y₃) - toh triangle ka area nikalna bahut straightforward hai. Formula hai: Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|. Par yeh kaise kaam karta hai?
Sochlo tum triangle ke boundary par walk karte ho: A se B, B se C, C se A. Har edge ke niche floor (x-axis) tak ek trapezoid banta hai. Trick yeh hai ki agar tum right side jaate ho toh trapezoid ko positive count karo, aur left side jaate ho toh negative. Yehi re
