Triangle abcabc has a perimeter of 1212 units. the vertices of the triangle are a(x, 2), b(2,−2)a(x, 2), b(2,−2), and c(−1, 2)c(−1, 2). find the value of xx.

Given:
The vertices of Δabc are
a: (x,2)
b: (2, -2)
c: (-1, 2)
The perimeter is 12 units.

Note that the distance between two points (x₁, y₁) and (x₂, y₂) is
d = √[(x₂ -x₁)² + (y₂ - y₁)²]

Let  d₁ =  distance between a and b
       d₂ =  distance between b and c
       d₃ =  distance between a and c
Then
  d₁ + d₂ + d₃ = 12

Calculate d₁²
d₁² = (x-2)² + 16
d₂² = 9 + 16 = 25
d₃² = (x+1)² 

Therefore
√[(x-2)²+16] + 5 + (x+1) = 12
√(x² - 4x + 4 + 16) = 12 - 5 - (x+1) = 6 - x
x² - 4x + 20 = 36 - 12x + x²
8x = 16
x = 2

Check the
a (2,2)
b (2, -2)
c (-1,2)
d₁ = 4
d₂ = 5
d₃ = 3
d₁ + d₂ +d₃ = 4 + 5 + 3 = 12  (Correct)

 x = 2

Main idea on left, details on right

I’m pretty sure the answer is a