Coordinate Geometry: Area of Triangle and Collinearity Class 10

In elementary geometry, calculating the area of a triangle is simple when the base and perpendicular height are known: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$. But what if a surveyor or civil engineer only has the GPS coordinates of three boundary markers on a map: $A(x_1, y_1), B(x_2, y_2),$ and $C(x_3, y_3)$? Calculating perpendicular heights using slopes and line equations would be excruciatingly tedious.

In CBSE Class 10 Mathematics, Chapter 7 (Coordinate Geometry) provides an elegant, direct coordinate area formula. Furthermore, this formula provides the single fastest test for collinearity: if the area of the triangle formed by three points equals zero, the points must lie on the exact same straight line!

In this master guide, we break down the area formula, solve for unknown parameters like $k$, and compute the area of polygons.

What You Will Learn

• Derivation and structure of the Area of a Triangle Formula in coordinate geometry
• The cyclical memory pattern: $1(2 - 3) + 2(3 - 1) + 3(1 - 2)$
• Why the absolute value modulus ($| \dots |$) is mandatory
• The Collinearity Condition: $\text{Area of } \Delta ABC = 0$
• Finding the value of $k$ when three points are given as collinear (CBSE 3-mark classic)
• Calculating the Area of a Quadrilateral by diagonal partitioning
• Solved board examination numerical problems and common traps

1. The Area of a Triangle Formula

For any triangle $\Delta ABC$ with vertices $A(x_1, y_1), B(x_2, y_2),$ and $C(x_3, y_3)$:

The Master Formula

$\mathbf{\text{Area}(\Delta ABC) = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|}$

                            Cyclic Index Pattern (1 -> 2 -> 3)
                                          1
                                        /                                          v                                           3 <----- 2
             x1 multiplies (y2 - y3)  ───> Indices follow 1, 2, 3
             x2 multiplies (y3 - y1)  ───> Indices follow 2, 3, 1
             x3 multiplies (y1 - y2)  ───> Indices follow 3, 1, 2

Important: <u>Why is the modulus sign ($| \dots |$) used? Because geometric area is a physical scalar quantity and can NEVER be negative! If your algebraic calculation inside the brackets yields $-15$, the physical area is $+15 ext{ sq. units}$.</u>

2. Condition for Collinearity of Three Points

Three points $A, B,$ and $C$ lie on a straight line (are collinear) if and only if the triangle formed by them has zero area!

The Collinearity Formula:

$\mathbf{\text{Area}(\Delta ABC) = 0}$ $\mathbf{x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0}$

(Notice the $\frac{1}{2}$ drops away because $\frac{1}{2} \times (\dots) = 0 \implies (\dots) = 0$).

3. High-Yield Solved Board Examination Problems

Solved Example 1: Finding $k$ for Collinear Points (NCERT Classic)

Problem: Find the value of $k$ for which the points $A(7, -2), B(5, 1),$ and $C(3, k)$ are collinear.

Solution:

1. List the Coordinates:
   • $(x_1, y_1) = (7, -2)$
   • $(x_2, y_2) = (5, 1)$
   • $(x_3, y_3) = (3, k)$
2. Apply the Collinearity Condition: $x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0$
3. Substitute the Values: $7(1 - k) + 5[k - (-2)] + 3(-2 - 1) = 0$ $7(1 - k) + 5(k + 2) + 3(-3) = 0$
4. Expand and Simplify: $7 - 7k + 5k + 10 - 9 = 0$ $(7 + 10 - 9) + (-7k + 5k) = 0$ $8 - 2k = 0$ $2k = 8 \implies \mathbf{k = 4}$
5. Therefore, <u>the value of $k$ for which the points are collinear is $4$</u>.

Solved Example 2: Area of a Quadrilateral (4-Mark Classic)

Problem: Find the area of the quadrilateral whose vertices, taken in order, are $A(-4, -2), B(-3, -5), C(3, -2),$ and $D(2, 3)$.

                         A (-4, -2) ------------ D (2, 3)
                         |       \               |
                         |        \  Diagonal    |
                         |         \   AC        |
                         B (-3, -5) ------------ C (3, -2)
                         Total Area = Area(ΔABC) + Area(ΔACD)

Solution:

1. Divide quadrilateral $ABCD$ into two triangles by drawing diagonal $AC$: $\mathbf{\text{Area}(ABCD) = \text{Area}(\Delta ABC) + \text{Area}(\Delta ACD)}$

2. Calculate Area of $\Delta ABC$:

   • Vertices: $A(-4, -2), B(-3, -5), C(3, -2)$ $\text{Area}(\Delta ABC) = \frac{1}{2} |(-4)[-5 - (-2)] + (-3)[-2 - (-2)] + 3[-2 - (-5)]|$ $\text{Area}(\Delta ABC) = \frac{1}{2} |(-4)(-3) + (-3)(0) + 3(3)|$ $\text{Area}(\Delta ABC) = \frac{1}{2} |12 + 0 + 9| = \frac{1}{2} |21| = \mathbf{\frac{21}{2}\text{ sq. units}}$

3. Calculate Area of $\Delta ACD$:

   • Vertices: $A(-4, -2), C(3, -2), D(2, 3)$ $\text{Area}(\Delta ACD) = \frac{1}{2} |(-4)[-2 - 3] + 3[3 - (-2)] + 2[-2 - (-2)]|$ $\text{Area}(\Delta ACD) = \frac{1}{2} |(-4)(-5) + 3(5) + 2(0)|$ $\text{Area}(\Delta ACD) = \frac{1}{2} |20 + 15 + 0| = \frac{1}{2} |35| = \mathbf{\frac{35}{2}\text{ sq. units}}$

4. Add the Two Areas: $\text{Area}(ABCD) = \frac{21}{2} + \frac{35}{2} = \frac{56}{2} = \mathbf{28\text{ sq. units}}$

5. Therefore, <u>the area of the quadrilateral is $28\text{ square units}$</u>.

4. Summary and Examination Tips

Exam Tip: When calculating the area of a quadrilateral, ensure the vertices are taken in order ($A \to B \to C \to D$). If you take them out of order, diagonal $AC$ will not divide the quadrilateral correctly!

Common Mistake: Subtraction sign errors with negative coordinates. In $[-2 - (-5)]$, writing $-2 - 5 = -7$ instead of $-2 + 5 = +3$. Always use parentheses for negative numbers!