Question

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

Answer 1

Answer:

The correct option is;

b) 0 sq, units

Step-by-step 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,  $A = \sqrt{s\cdot \left (s-a \right )\cdot \left (s-b \right ) \cdot \left ( s-c \right )}$

Where;

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

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

Therefore, the area = 0 sq, units.