Question

Using integration find the area of triangle whose vertices are (1,1) (3,4) and (5,0)

Answer 1

Let the given vertices respectively be A,B and C
Equation of line AB is: y=(3/2)x-1/2
Equation of line BC is: y= - 2x+10
Equation of line AC is: y=-(1/4)x+5/4
Area of required triangle will therefore be= area under AB(with limits x=1 to x=3) + Area under BC(with limits x=3 to x=5) + Area under AC (with limits x+5 to x=1) = $\int\limits^3_1 {[(3/2)x-1/2]} \, dx + \int\limits^5_3 {[- 2x+10]} \, dx + \int\limits^1_5 {[-(1/4)x+5/4]x} \, dx$
Solving this you get the area under the required triangle.