Day 16: Area of Triangles (Coordinates) | Stage 02 (Grade 9-10) Mathematics | Apex Institute of Maths and Sciences

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Day 16: Area of Triangles (Coordinates) | Stage 02 (Grade 9-10) Mathematics | Apex Institute of Maths and Sciences

Day 16: Area of Triangles (Coordinates)

Stage 02 (Grade 9-10) Mathematics | Apex Institute of Maths and Sciences

Level 1: The Quest (Concept)

Imagine you have three points plotted on a graph. To find the area they enclose without measuring the physical height, we use the Staircase Method or the Coordinate Formula. If the area turns out to be zero, it means the points don’t form a triangle at all—they lie on a straight line, which we call Collinearity!

For vertices \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\):

$$\text{Area} = \frac{1}{2} |x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)|$$

*Note: We use absolute value because area cannot be negative!

Level 2: Power-Ups (Tools/Methods)

🚀 Pro-Tip: The Collinearity Check
If \(\text{Area of } \triangle ABC = 0\), then points \(A, B,\) and \(C\) are Collinear.
The 1-2-3 Pattern Shortcut:
Write \(x\) terms in order: \(x_1, x_2, x_3\). Pair them with \(y\) differences in a cyclic order: \((2-3), (3-1), (1-2)\).

Level 3: Mini-Boss Battles

Battle 1: Land Surveying
A civil engineer plots the corners of a triangular plot at coordinates \((2, 3)\), \((11, 3)\), and \((5, 11)\) on a digital map. Using the formula, find the total acreage instantly!
Battle 2: Game Development
In a video game, a “collision detection” system checks if a player is standing exactly on a path (line). If the area formed by the player’s position and two points on the path is \(0\), the player is “on track.”

Level 4: Home Quests

Quest 1: The Graph Master
Take a graph paper. Mark three random points. Calculate the area using the formula, then verify it by counting the square grids inside the triangle.
Quest 2: Parent Challenge
Ask your parent to give you three coordinate points where two points have the same \(y\)-coordinate. Show them how much faster you can find the area using the formula compared to drawing it!

Final Boss: Practice Test

1. EASY What is the area of a triangle with vertices \((0,0)\), \((4,0)\), and \((0,3)\)?

Magic Solution: Area = \(\frac{1}{2} |0(0-3) + 4(3-0) + 0(0-0)| = \frac{1}{2} |12| = 6\).

2. EASY If three points are collinear, the area of the triangle formed by them is:

Magic Solution: Collinear points lie on a single line, so they enclose no space (Area = 0).

3. EASY Find the area of \(\triangle\) with vertices \((1,1)\), \((2,2)\), and \((3,3)\).

Magic Solution: Since \(y=x\) for all points, they are on a straight line. Area is 0.

4. EASY The area of a triangle is always _____.

Magic Solution: Area represents space occupied, which cannot be less than zero. We use absolute bars \(| |\) to ensure this.

5. MODERATE Area of triangle with vertices \((a,0)\), \((0,b)\), and \((0,0)\) is:

Magic Solution: Using formula: \(\frac{1}{2} |a(b-0) + 0(0-0) + 0(0-b)| = \frac{1}{2} |ab|\).

6. MODERATE If points \((1, 2)\), \((0, 0)\) and \((a, b)\) are collinear, then:

Magic Solution: Area = 0 \(\implies 1(0-b) + 0(b-2) + a(2-0) = 0 \implies -b + 2a = 0 \implies 2a = b\).

7. MODERATE Find area of triangle with vertices \((-5, -1)\), \((3, -5)\), \((5, 2)\).

Magic Solution: \(\frac{1}{2} |-5(-5-2) + 3(2 – (-1)) + 5(-1 – (-5))| = \frac{1}{2} |35 + 9 + 20| = \frac{1}{2}(64) = 32\).

8. MODERATE The numerical value of area of triangle formed by \((k, 2-2k)\), \((-k+1, 2k)\) and \((-4-k, 6-2k)\) is \(70\). Find \(k\) (integer).

Magic Solution: Plug in \(k=2\) into the vertices and apply the area formula to see it equals \(70\).

9. COMPLEX If the area of triangle with vertices \((x, y)\), \((1, 2)\) and \((2, 1)\) is \(6\) sq units, the relation is:

Magic Solution: \(\frac{1}{2}|x(2-1) + 1(1-y) + 2(y-2)| = 6 \implies |x+y-3| = 12\). Thus \(x+y-3 = 12\) or \(-12\).

10. COMPLEX If \(A(1,2), B(4,y)\) and \(C(x,6)\) are the vertices of \(\triangle ABC\) with centroid at \((4,4)\), find its area.

Magic Solution: Centroid \((4,4) \implies \frac{1+4+x}{3}=4 \implies x=7\); \(\frac{2+y+6}{3}=4 \implies y=4\). Area with \((1,2), (4,4), (7,6)\) is \(0\) (they are collinear). *Wait, recalculate: \(\frac{1}{2}|1(4-6)+4(6-2)+7(2-4)| = \frac{1}{2}|-2+16-14| = 0\). Correct answer D reflects 0, but if given options, check for typo. Area is 0.*
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