Question

Use definite integration, find the area of triangle whose vertices are (-1,6),(1,2) and (5,4).

Answer 1

The area of triangle is 29 square unit.

Step-by-step explanation:

To find : The area of triangle formed by vertices (-1,6),(1,2) and (5,4) ?

Solution :

The area of the triangle is given by,

$A=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Here, $(x_1,y_1)=(-1,6),(x_2,y_2)=(1,2),(x_3,y_3)=(5,4)$

$A=\frac{1}{2}[-1(2-4))+1(4-6)+5(6-2)]$

$A=\frac{1}{2}[-1(-2))+1(-2)+5(4)]$

$A=\frac{1}{2}[2-2+20]$

$A=\frac{1}{2}[20]$

$A=10$

Therefore, the area of triangle is 10 square unit.

#Learn more

What is the area of the triangle with the vertices

https://brainly.in/question/1719632

Answer 2

The area of triangle is 29 square unit.

Step-by-step explanation:

To find : The area of triangle formed by vertices (-1,6),(1,2) and (5,4) ?

Solution :

The area of the triangle is given by,

$A=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Here, $(x_1,y_1)=(-1,6),(x_2,y_2)=(1,2),(x_3,y_3)=(5,4)$

$A=\frac{1}{2}[-1(2-4))+1(4-6)+5(6-2)]$

$A=\frac{1}{2}[-1(-2))+1(-2)+5(4)]$

$A=\frac{1}{2}[2-2+20]$

$A=\frac{1}{2}[20]$

$A=10$