Answer:
The value of x is 2
Step-by-step explanation:
* Lets explain how to find a distance between 2 points
- If the endpoints of a segment are [tex](x_{1},y_{1})[/tex] and
[tex](x_{2},y_{2})[/tex] is [tex]d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]
- Triangle ABC has a perimeter of 12 units
∵ The perimeter of any triangle is the sum of lengths of its sides
∴ P Δ ABC = AB + BC + AC
* Lets find the length of the three sides
∵ A = (x , 2) , B = (2 , -2) , C = (-1 , 2)
∵ [tex]AB=\sqrt{(2-x)^{2}+(-2-2)^{2}}[/tex]
∴ [tex]AB=\sqrt{(2-x)^{2}+(-4)^{2}}[/tex]
∴ [tex]AB=\sqrt{(2-x)^{2}+16}[/tex]
∵ [tex]BC=\sqrt{(-1-2)^{2}+(2--2)^{2}}[/tex]
∴ [tex]BC=\sqrt{(-3)^{2}+(4)^{2}}[/tex]
∴ [tex]BC=\sqrt{9+16}[/tex]
∴ [tex]BC=\sqrt{25}[/tex]
∴ BC = 5
∵ [tex]CA=\sqrt{(x--1)^{2}+(2-2)^{2}}[/tex]
∴ [tex]CA=\sqrt{(x+1)^{2}+(0)^{2}}[/tex]
∴ [tex]CA=\sqrt{(x+1)^{2}}[/tex]
- The √ is canceled by power 2
∴ CA = (x + 1)
∵ AB + BC + CA = 12
∴ [tex]\sqrt{(2-x)^{2}+16}[/tex] + 5 + (x + 1) = 12
- Add 5 and 1
∴ [tex]\sqrt{(2-x)^{2}+16}[/tex] + 6 + x = 12
- subtract 6 and x from both sides
∴ [tex]\sqrt{(2-x)^{2}+16}[/tex] = (6 - x)
- To cancel (√ ) square the two sides
∴ (2 - x)² + 16 = (6 - x)²
- Simplify the two sides
∴ [(2)(2) + (2)(2)(-x) + (-x)(-x)] + 16 = (6)(6) + (2)(6)(-x) + (-x)(-x)
∴ 4 - 4x + x² + 16 = 36 - 12x + x²
- Subtract x² from both sides
∴ 20 - 4x = 36 - 12x
- Add 12x to both sides and subtract 20 from both sides
∴ 12x - 4x = 36 - 20
∴ 8x = 16
- Divide both sides by 8
∴ x = 2
* The value of x is 2
