Question

Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices area A(1, 2), B (2, 0) and C (4, 3).

Answer 1

Let A(1, 2), B (2, 0) and C (4, 3) be the vertices of a triangle ABC

$$\text{ Area of }△ABC=\text{ Area of trapezium }ADEC-\text{ Area of }△ADB-\text{ Area of }△CBE$$

$$\text{ Equation of sides AC, AB and BC are given by }:$$

$$y=\frac{x}{3}+\frac{5}{3},y=-2x+4\text{ and }y=\frac{3x}{2}-3\text{P respectively }$$

$$\text{ Hence, area of }△ABC={\int }_1^4(\frac{x}{3}+\frac{5}{3})dx-{\int }_1^2(-2x+4)dx-{\int }_2^4(\frac{3}{2}x-3)dx$$

$$=\frac{1}{3}{[\frac{x^2}{2}+5x]}_1^4-{[-\frac{2x^2}{2}+4x]}_1^2-{[\frac{3}{4}{x}^2-3x]}_2^4$$

$$=\frac{1}{3}[(\frac{16}{2}+20)-(\frac{1}{2}+5)]-[(-4+8)-(-1+4)]-[(12-12)-(3-6)]$$

$$=\frac{1}{3}[28-\frac{11}{2}]-[4-3]-[0+3]$$

$$=\frac{1}{3}\left[\frac{45}{2}\right]-\left[1\right]-\left[3\right]$$

$$=\frac{7}{2}$$