#### Question

Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices are A (4 , 1), B (6, 6) and C (8, 4).

#### Solution

Equation of AB

`y - y_1 = (y_2 -y_1)/(x_2-x_1) (x - x_1)`

`y - 1 = (6-1)/(6-4) (x - 4)`

`y - 1 = 5/2 (x - 4)`

2y - 2 = 5x - 20

`y = (5x)/2 - 9`

Equation of BC

`y - 6 = (4 - 6)/(8 - 6) (x - 6)`

`y - 6 = (-2)/(+2) (x - 6)`

y - 6 = -x + 6

y = -x + 12

Equation of AC

`y - 1 = (4 -1)/(8 - 4) (x - 4)`

`y -1 = 3/4 (x - 4)`

`4y - 4 = 3x - 12`

`y = (3x)/4 - 2`

Is there an error in this question or solution?

Solution Using the Method of Integration, Find the Area of the Triangle Abc, Coordinates of Whose Vertices Are a (4 , 1), B (6, 6) and C (8, 4). Concept: Area Under Simple Curves.