Find the area of the triangle whose vertices are (2, 3); (-1, 0); (2, -4)... [Also show the steps of the solution]
Please answer correctly, it's really urgent!​

Area = 14.5

Step-by-step explanation:

Let's label the given coordinates A, B, C;

Thus;

A = (2, 3)

B = (-1, 0)

C = (2, -4)

From the coordinate geometry formula, the formula for area of a triangle with 3 vertices is;

Area = [Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By)]/2

Area = [2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - (-4))]/2

Area = 29/2

Area = 14.5

answer: #1 ads up by 8 and i don't know #2

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct