Question

)If the vertices of a triangle are (3,1);(4,2); and(5,3). Then area is: *​

Answer 1

Answer:

0 sq. units

Step-by-step explanation:

Given:

Vertices of the triangle = (3, 1), (4, 2), (5, 3)

To find:

The area of the triangle

Solution:

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

Here, $x_1=3, x_2=4, x_3=5, y_1=1, y_2=2, y_3=3$

So, the area of the triangle = $\frac{1}{2}[3(2-3)+4(3-1)+5(1-2)]$

⇒ $\frac{1}{2}[3(2-3)+4(3-1)+5(1-2)]$

⇒ $\frac{1}{2}[3(-1)+4(2)+5(-1)]$

⇒ $\frac{1}{2}[-3+8-5]$

⇒ $\frac{1}{2}[8-8]$

⇒ $\frac{1}{2}\times 0$

⇒ $0$ sq. unit

∴ The area of the triangle is $0$ sq. units.