NCERT · Class 10 · Mathematics · Coordinate GeometryFind the area of a triangle whose vertices are $(1, -1), (-4, 6)$ and $(-3, -5)$. Also, derive and explain the formula used to calculate the area of a triangle in coordinate geometry.

Introduction to Area of a Triangle in Coordinate Geometry\nIn elementary geometry, the area of a triangle is typically calculated using the formula $\frac{1}{2} \times \text{base} \times \text{height}$. However, when only the Cartesian coordinates of the three vertices are provided, determining the height can be tedious. Coordinate geometry provides a direct formula derived from trapezoids or smaller triangles to calculate the area directly from coordinates.

Derivation of the Area Formula

1. Plotting Vertices: Let $A(x_1, y_1), B(x_2, y_2),$ and $C(x_3, y_3)$ be the vertices of a triangle $\triangle ABC$.
2. Drawing Perpendiculars: Draw perpendiculars $AP, BQ,$ and $CR$ from the vertices $A, B,$ and $C$ respectively onto the $X$-axis (with $P, Q, R$ on the $X$-axis).
3. Forming Trapezoids: This construction creates three trapezoids: $ABQP, APRC,$ and $BQRC$.
4. Relating Areas: The area of $\triangle ABC$ can be geometrically expressed by combining the areas of trapezoids: $$\text{Area}(\triangle ABC) = \text{Area of trapezoid } ABQP + \text{Area of trapezoid } APRC - \text{Area of trapezoid } BQRC$$
5. Applying Trapezoid Area Formula ($\frac{1}{2} \times \text{sum of parallel sides} \times \text{height}$): $$\text{Area} = \frac{1}{2}(y_1 + y_2)(x_1 - x_2) + \frac{1}{2}(y_1 + y_3)(x_3 - x_1) - \frac{1}{2}(y_2 + y_3)(x_3 - x_2)$$
6. Simplifying the Expression: Expanding and grouping terms yields the standard determinant-based coordinate formula: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Numerical Step-by-Step Solution\nGiven vertices: $A(x_1, y_1) = (1, -1)$, $B(x_2, y_2) = (-4, 6)$, and $C(x_3, y_3) = (-3, -5)$.

• Step 1: Write the area formula: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

• Step 2: Substitute the coordinate values into the formula: $$\text{Area} = \frac{1}{2} |1(6 - (-5)) + (-4)(-5 - (-1)) + (-3)((-1) - 6)|$$

• Step 3: Simplify the inner terms:

  • $1(6 + 5) = 1(11) = 11$
  • $(-4)(-5 + 1) = (-4)(-4) = 16$
  • $(-3)(-7) = 21$

• Step 4: Sum the simplified values: $$\text{Area} = \frac{1}{2} |11 + 16 + 21|$$ $$\text{Area} = \frac{1}{2} |48|$$ $$\text{Area} = 24 \text{ square units}$|

Conclusion\nThe area of the triangle formed by the given vertices is $24$ square units.