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

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;

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,  

Where;

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

Therefore, the area = 0 sq, units.

y < -3

step-by-step explanation:

-3y - 2 > 7

-3y -2 + 2 > 7+2

-3y > 9

-3y/-3 < 9/-3   (change the inequality sign when multiplying or dividing by a negative number)

y < -3