Q:

Triangle ABC has a perimeter of 12 units. The vertices of the triangle are A(x,2), B(2,-2), and C(-1,2). Find the value of x.

Accepted Solution

A:
Answer:The value of x is 2Step-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