Browse Questions

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

Toolbox:
• Equation of a line when two points are given is $\large\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}.$
Step 1:
Let the points be $A(1,3),B(2,5),C(3,4)$
The equation of the line AB is $\large\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}.$
(i.e) $\large\frac{y-3}{5-3}=\frac{x-1}{2-1}.$
$\Rightarrow y-3=2x-2$
$2x-y=-1$
$\Rightarrow y=2x+1$-----(1)
The equation of the line BC is
(i.e) $\large\frac{y-5}{4-5}=\frac{x-2}{3-2}.$
$\Rightarrow y-5=-1(x-2)$
$x-y=7$
$\Rightarrow y=7-x$------(2)
The equation of the line AC is
(i.e) $\large\frac{y-3}{4-3}=\frac{x-1}{3-1}.$
$\Rightarrow2( y-3)=x-1$
$x-2y=-5$
$\large\frac{x+5}{2}$$=y----(3) Step 2: Area of the required region is A=\int_1^2y_1+\int_2^3y_2-\int_1^3y_3 \;\;=\int_1^2(2x+1)dx+\int_2^3(7-x)-\large\frac{1}{2}$$\int_1^3(x+5)dx$
On integrating we get,
$\big[\large\frac{2x^2}{2}$$+x\big]_1^2+\big[7x-\large\frac{x^2}{2}$$+x\big]_2^3-\large\frac{1}{2}\big[\large\frac{x^2}{2}$$+5x\big]_1^3 Step 3: On applying limits we get, (4+2)-(1+1)+(21-\large\frac{9}{2})$$-(14-\large\frac{4}{2})-\frac{1}{2}\big[(\large\frac{9}{2}$$+15)-(\large\frac{1}{2}$$+5)]$
On simplifying we get,
$A=\large\frac{3}{2}$sq.units