To find the perimeter of the triangle formed by the points \((-4, 2)\), \( (1, 2)\), and \( (1, 6)\), we will first compute the lengths of the sides of the triangle using the distance formula. The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Step 1: Calculate the lengths of each side.
1. Side between \((-4, 2)\) and \( (1, 2)\):
\[ d_1 = \sqrt{(1 - (-4))^2 + (2 - 2)^2} \] \[ d_1 = \sqrt{(1 + 4)^2 + 0^2} \] \[ d_1 = \sqrt{5^2} = 5 \]
2. Side between \( (1, 2)\) and \( (1, 6)\):
\[ d_2 = \sqrt{(1 - 1)^2 + (6 - 2)^2} \] \[ d_2 = \sqrt{0^2 + (6 - 2)^2} \] \[ d_2 = \sqrt{4^2} = 4 \]
3. Side between \( (-4, 2)\) and \( (1, 6)\):
\[ d_3 = \sqrt{(1 - (-4))^2 + (6 - 2)^2} \] \[ d_3 = \sqrt{(1 + 4)^2 + (6 - 2)^2} \] \[ d_3 = \sqrt{5^2 + 4^2} \] \[ d_3 = \sqrt{25 + 16} = \sqrt{41} \]
Step 2: Compute the perimeter.
Now we can find the perimeter \(P\) of the triangle by adding the lengths of the sides:
\[ P = d_1 + d_2 + d_3 = 5 + 4 + \sqrt{41} \]
We can approximate \(\sqrt{41}\):
\[ \sqrt{41} \approx 6.4 \quad (\text{since } 6.4^2 = 40.96 \text{ is close to } 41) \]
Therefore, substituting this back into the perimeter equation:
\[ P \approx 5 + 4 + 6.4 = 15.4 \]
Step 3: Round to the nearest tenth.
The perimeter of the triangle is approximately:
\[ \boxed{15.4} \]