Atriangle with vertices located at (−2, −3) and (3, −3) has an area of 20 square units. determine one possible location of the other vertex.

Both given points have the same Y coordinate, so this can be the base of the triangle. The length of the base id the sum of the X values:

-3 to 3 is 5 units long.

Area of a triangle is 1/2 x base x height.

The area is given as 20 square units.

Calculate the height:

20 = 1/2 x 5 x Height

20 = 2.5 x height

Height = 20/2.5

Height = 8 units.

Using the point (-2,-3) keep the x value -2 then add 8 units to the Y value for the height:

The 3rd vertex could be (-2,5)

for finding the midpoint of any given 2 pair of points simply add the respective x coordinates with other x coordinate and the y coordinate with other y coordinate and divide them by 2

in your question it will be :

(3+1)/2 for the x coordinate of mid point i.e 2

[8+(-4)]/2 for the y coordinate of midpoint i.e 2

so coordinates of mid point of points (3,8) and (1,-4) will be (2,2)

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180