Question

Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B(4, 5) and C(6, 3).

Answer 1

The given points are A (2, 0), B(4, 5) and C(6, 3) Equation of line AB is
$y-0=\frac{5-0}{4-2}(X-2)$ [Using $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)]$
$\Rightarrow y=\frac{5}{2}x-5$
Equation of line BC is
$y-0=\frac{3-0}{6-2}(x-2)\Rightarrow y=\frac{3}{4}x-\frac{3}{2}$
Required area = (Area under line segment AB) + (Area under line
segment BC) - (Area under line segment AC)
$={\int }_5^4(\frac{5}{2}x-5)dx+{\int }_4^6(-x+9)dx-{\int }_2^6(\frac{3}{4}x-\frac{3}{2})dx=[\frac{5x^2}{4}-5x{]}_2^4+[\frac{-x^2}{2}+9x{]}_4^6-[\frac{3x^2}{8}-\frac{3}{2}x{]}_2^6=(\frac{5\times 4^2}{4}-5\times 4)-(\frac{5\times 2^2}{4}-5\times 2)+(\frac{-6^2}{2}+9\times 6)-(\frac{-4^2}{2}+9\times 4)-(\frac{3\times 6^2}{8}-\frac{3}{2}\times 6)+(\frac{3\times 2^2}{8}-\frac{3}{2}\times 2)=(20-20)-(5-10)+(-18+54)-(-8+36)-(\frac{27}{2}-9)+(\frac{3}{2}-3)=536-28-\frac{9}{2}-\frac{3}{2}=7squnit$