Question

Find the area of the triangle whose vertices are (4 , 7) (1 , 3) (5 , 1)

Answer 1

ans is 11unit^2. you can solve it by determinant method or
1/2|4×3 +1×1 +5×7 -1×7-5×3-4×1|
=11 unit^2
OR
BY DETERMINANT METHOD
4 7 1
1 3 1
5 1 1

On solving it
1/2[4(3-1) -7(1-5) +1(1-15)]
=11 unit^2
hope it helps u

Answer 2

Answer:

The area of the triangle = 11 square units

Step-by-step explanation:

Given data

Vertices of triangle are (4, 7) (1,3) and (5,1)

area of the triangle with vertices ($x_{1} , y_{1}$ ) ($x_{2} , y_{2}$) ($x_{3} , y_{3}$) is given by

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

⇒ the area of the triangle with given vertices

            =  $\frac{1}{2}$ | 4 (3-1) + 1 (1-7) +5 (7-3) |  

            = $\frac{1}{2}$ | 4(2) +(-6)+5(4) |  

            = $\frac{1}{2}$ | 8 - 6 +20 |  

            = $\frac{1}{2}$ | 2+ 20|

            = $\frac{1}{2}$ (22) = 11  square units