Question

1. If the area of a triangle with vertices (5,2), (x,4), and (0,-3) is 25/2 sq.
units. Find x.
2. If the area of a triangle with vertices (-2,0), (4,0), and (3,y) is 9 sq.
units. Find y.
pls po pasagot sana salamat

Answer 1

Answer:

The value of x = 3 and the value of y = 3

Step-by-step explanation:

Given data  

1) area of the triangle with vertices (5,2), (x,4), and (0,-3) is 25/2 sq units

2) area of the triangle with vertices (-2,0), (4,0), and (3,y) is 9 sq units

Here we need to find x and y values  

⇒ Area of the triangle with vertices  (x₁ ,y₁)   (x₂, y₂)   (x₃, y₃)  

              = $\frac{1}{2}$ [ x₁ (y₂-y₃) +x₂ (y₃-y₁) +x₃(y₂-y₁)]

⇒ area of the triangle with vertices (5,2), (x,4), and (0,-3) is 25/2 sq units  

      ⇒ $\frac{1}{2}$ [ 5(4-(-3)) + x(-3-2) + 0(4-2)] = 25/2

      ⇒ 1/2 [ 5(7) + x(-5) + 0] = 25/2    

      ⇒ 35 -5x =25  

      ⇒  - 5x = -15

      ⇒  x =3

⇒ area of the triangle with vertices (-2,0), (4,0), and (3,y) is 9 sq units  

      ⇒ $\frac{1}{2}$ [ -2(0-y) + 4(y-0)+ 3 (0-0) ] = 9

      ⇒ $\frac{1}{2}$ [ -2(-y) + 4y + 0] = 9

      ⇒     2y + 4y  = 18

      ⇒     6y = 18

      ⇒       y = 3

⇒ The value of x = 3 and y = 3