Question

find the area of triangle whose vertices are (x1,y1),(x2,y2)and(x3,y3)​

Answer 1

Answer:

There is a formula by which you can calculate the area of a triangle whose coordinates of its vertices are given.

Area= 1/2 |x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|    

We have taken modulus here as the value of area shouldn't be negative.

Answer 2

The area of triangle is $\dfrac{1}{2}[x_{1}(y_{2}-y_{1})+x_{2}(y_{3}-y_{2})+x_{3}(y_{1}-y_{2})]$.

Step-by-step explanation:

Given,

The three vertices of triangle are $(x_1,y_1), (x_2,y_2)$ and $(x_3, y_3)​.$.

To find, the area of triangle whose vertices are $(x_1,y_1), (x_2,y_2)$ and $(x_3, y_3)​.$ =?

We know that,

The area of triangle

= $\dfrac{1}{2}[x_{1}(y_{2}-y_{1})+x_{2}(y_{3}-y_{2})+x_{3}(y_{1}-y_{2})]$

∴ The area of triangle whose vertices are $(x_1,y_1), (x_2,y_2)$ and $(x_3, y_3)​.$

= $\dfrac{1}{2}[x_{1}(y_{2}-y_{1})+x_{2}(y_{3}-y_{2})+x_{3}(y_{1}-y_{2})]$

Thus, the area of triangle is $\dfrac{1}{2}[x_{1}(y_{2}-y_{1})+x_{2}(y_{3}-y_{2})+x_{3}(y_{1}-y_{2})]$.