Question

Using the method of integration, the area of the triangular region whose vertices are $A(2,-2),B(4,3),C(1,2)$ comes out to be $\frac{k}{2}$, find $k$

Answer 1

Equation of line $AB:$

$y+2=\left(\frac{3+2}{2}\right)(x-2)$

$\Rightarrow 2y=5x-14$

Equation of line BC:

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

$\Rightarrow 3y=x+5$

Equation of line CA:

$y-2=-4(x-1)$

$\Rightarrow 4x+y=6$

Therefore, area of triangle ABC, = ${\int }_{-2}^3\frac{2y+14}{5}dy-{\int }_2^3(3y-5)dy-{\int }_{-2}^2\frac{6-y}{4}dy$

$=\frac{75}{5}-\frac{5}{2}-\frac{24}{4}$

$=\frac{300-120-50}{20}=\frac{130}{20}$

Area $=\frac{13}{2}$ sq. units.

So $\frac{k}{2}=\frac{13}{2}$

$\Rightarrow k=13$