Question

On a coordinate plane, triangle A B C has points (1, 1), (4, 0), (3, 5).What is the area of triangle ABC?

Answer 1

Answer:

Area of Triangle ABC = $\frac{13\sqrt{858} }{4}$  unit² or 7.32 unit²

Step-by-step explanation:

Given coordinates of Triangle are

A = ( 1 , 1)

B = ( 4 , 0)

C = ( 3 , 5)

So , The measure of side AB is

AB = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

or, AB =  $\sqrt{(4-1)^{2}+(0-1)^{2}}$

Or, AB =  $\sqrt{(3)^{2}+(-1)^{2}}$

∴  AB = $\sqrt{10}$

Again ,

The measure of side BC is

BC = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

or, BC =  $\sqrt{(3-4)^{2}+(5-0)^{2}}$

Or, BC =  $\sqrt{(-1)^{2}+(5)^{2}}$

∴  BC = $\sqrt{26}$

Similarly ,

The measure of side CA is

CA = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

or, CA =  $\sqrt{(3-1)^{2}+(5-1)^{2}}$

Or, CA =  $\sqrt{(2)^{2}+(4)^{2}}$

∴  CA = $\sqrt{20}$

Now, let D be the mid points of side BC

So, Points ( D ) = $\frac{(x_1+x_2)}{2}$  ,  $\frac{(y_1+y_2)}{2}$

I.e points d =  $\frac{(4+3)}{2}$  ,  $\frac{(0+5)}{2}$

Or, points D = $\frac{(7)}{2}$ , $\frac{(5)}{2}$

Now Measure of line AD is

AD =  $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

AD = $(\sqrt{\frac{7}{2}-1})^{2}$ +  $(\sqrt{\frac{5}{2}-1})^{2}$

Or, AD =( $\sqrt{\frac{5}{2} }$ )² + ( $\sqrt{\frac{3}{2} }$ )²

Or, AD = $\sqrt{\frac{33}{4} }$

Now, Area of Triangle ABC = $\frac{1}{2}$ × length × base

or, Area of Triangle ABC = $\frac{1}{2}$ × AD × BC

or, Area of Triangle ABC = $\frac{1}{2}$ ×  $\sqrt{\frac{33}{4} }$ ×  $\sqrt{26}$

Or, Area of Triangle ABC = $\frac{13\sqrt{858} }{4}$ unit²   or , 7.32  unit² Answer

Answer 2

Answer:

7 square units.

Step-by-step explanation:

On the coordinate plane, triangle ABC has vertices at (1,1), (4,0) and (3,5).

So, the area of the triangle ABC is given by  

$\frac{1}{2} | 1(0 - 5) + 4(5 - 1) + 3(1 - 0) | = 7$ square units. (Answer)

We know the formula that area of triangle having vertices at $(x_{1}, y_{1})$, $(x_{2}, y_{2})$, and $(x_{3}, y_{3})$ is given by  

Δ = $\frac{1}{2} | x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})|$