Question

Using integration, find the area of the region bounded by the triangle whose vertices are $(-1,2),(1,5)$ and $(3,4).$

Answer 1

Let $A=(-1,2)$

$B=(1,5)$

$C=(3,4)$

We have to find the area of $\Delta ABC$,

Find equation of line AB $\to y-5=\left(\frac{2-5}{-1-1}\right)(x-1)$

$y-5=\frac{3}{2}(x-1)$

$2y-10=3x-3$

$3x-2y+7=0$ ---- (1)

$y=\frac{3x+7}{2}$

Equation of line BC $\to y-4=\left(\frac{5-4}{1-3}\right)(x-3)$

$y-4=\frac{1}{-2}(x-3)$

$2y-8=-x+3$

$x+2y-11=0$---(2)

$y=\frac{11-x}{2}$

Equation of line AC$\to y-4=\left(\frac{2-4}{-1-3}\right)(x-3)$

$y-4=\frac{1}{2}(x-3)$

$2y-8=x-3$

$x-2y+5=0$---(3)

$y=\frac{x+5}{2}$

So, the required area = ${\int }_{-1}^1\left(\frac{3x+7}{2}\right)dx+{\int }_1^3\left(\frac{11-x}{2}\right)dx-{\int }_{-1}^3\left(\frac{x+5}{2}\right)dx$

$=\frac{1}{2}{[\frac{3x^2}{2}+7x]}_{-1}^1+\frac{1}{2}{[11x-\frac{x^2}{2}]}_1^3-\frac{1}{2}{[\frac{x^2}{2}+5x]}_{-1}^3$

$=\frac{1}{2}[[(\frac{3}{2}+7)-(\frac{3}{2}-7)]+\frac{1}{2}[(33-\frac{9}{2})-(11-\frac{1}{2})]-\frac{1}{2}[(\frac{9}{2}+15)-(\frac{1}{2}-5)]]$

$=\frac{1}{2}[14+22-4-24]\Rightarrow \frac{1}{2}[36-28]=4$ sq.units.

Therefore, the area of the triangle is $4$ sq.units.