Question

What is the area of a triangle whose vertices are (4,0), (2,

3), (8,-6)?(using distance formula)
a) 2 sq. units b) 0 sq. units c) 1 sq. units d) 4 sq. units​

Answer 1

The correct option is;

b) 0 sq, units

Explanation:

The vertices of the triangle are;

(4, 0), (2, 3), (8, -6)

The distance formula fr finding the length of a segment is given as follows;

`l = \sqrt{\left (y_(2)-y_(1) \right )^(2)+\left (x_(2)-x_(1) \right )^(2)}`

Where, (x₁, y₁) and (x₂, y₂) are the coordinates of the end points of the line

For the points (4, 0) and (2, 3) , we have;

√((3 - 0)² + (2 -4)²) = √13

Distance from (4, 0) to (2, 3) = √13

For the points (4, 0) and (8, -6) , we have;

√((-6 - 0)² + (8 -4)²) = √13 =

Distance from (4, 0) to (8, -6) = 2·√13

For the points (2, 3) and (8, -6) , we have;

√((-6 - 3)² + (8 -2)²) = 3·√13 =

Distance from (2, 3) to (8, -6) = 3·√13

Therefore, the perimeter of the triangle = 6·√13

The semi perimeter s = 3·√13

The area of the triangle,
`A = √(s\cdot \left (s-a \right )\cdot \left (s-b \right ) \cdot \left ( s-c \right ))`

Where;

a, b, and c are the length of the sides of the triangle;

`A = \sqrt{3\cdot √(3) \cdot \left (3\cdot √(3) -√(3) \right )\cdot \left (3\cdot √(3) -2 \cdot √(3) \right ) \cdot \left ( 3\cdot √(3) -3\cdot √(3) \right )} = 0`

Therefore, the area = 0 sq, units.