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

1. Calculate the distance AB<br /> Use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ for points $A(x, 2)$ and $B(2, -2)$. <br /> $AB = \sqrt{(2 - x)^2 + (-2 - 2)^2} = \sqrt{(2 - x)^2 + 16}$.<br /><br />2. Calculate the distance BC<br /> Use the distance formula for points $B(2, -2)$ and $C(-1, 2)$.<br /> $BC = \sqrt{(-1 - 2)^2 + (2 + 2)^2} = \sqrt{9 + 16} = 5$.<br /><br />3. Calculate the distance CA<br /> Use the distance formula for points $C(-1, 2)$ and $A(x, 2)$.<br /> $CA = \sqrt{(x + 1)^2 + (2 - 2)^2} = |x + 1|$.<br /><br />4. Set up the perimeter equation<br /> The perimeter is given as 12 units. Therefore, $AB + BC + CA = 12$.<br /> $\sqrt{(2 - x)^2 + 16} + 5 + |x + 1| = 12$.<br /><br />5. Simplify and solve for x<br /> Simplify to find $|x + 1| = 7 - \sqrt{(2 - x)^2 + 16}$.<br /> Consider two cases for $|x + 1|$: $x + 1 = 7 - \sqrt{(2 - x)^2 + 16}$ and $x + 1 = -(7 - \sqrt{(2 - x)^2 + 16})$.<br /> Solve each case separately.<br /><br />6. Solve Case 1: $x + 1 = 7 - \sqrt{(2 - x)^2 + 16}$<br /> Rearrange to get $\sqrt{(2 - x)^2 + 16} = 6 - x$.<br /> Square both sides: $(2 - x)^2 + 16 = (6 - x)^2$.<br /> Expand and simplify: $4 - 4x + x^2 + 16 = 36 - 12x + x^2$.<br /> Simplify further: $20 - 4x = 36 - 12x$.<br /> Solve for $x$: $8x = 16$, so $x = 2$.<br /><br />7. Verify solution<br /> Check if $x = 2$ satisfies the original perimeter condition.<br /> If $x = 2$, then $AB = \sqrt{0 + 16} = 4$, $BC = 5$, $CA = |2 + 1| = 3$.<br /> Perimeter $= 4 + 5 + 3 = 12$, which matches the given perimeter.