Question \( 2(1 \) point) \( \begin{array}{ll}\text { 11) }(-10,-7),(-8,1) & \text { 12) }(-6,-10),(-2,-10) \\ \text { Find the distance (Round to tenths place) }\end{array} \)

To find the distance between the points \((-10, -7)\) and \((-8, 1)\), we can use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Here, \((x_1, y_1) = (-10, -7)\) and \((x_2, y_2) = (-8, 1)\). Calculating: 1. \(x_2 - x_1 = -8 - (-10) = 2\) 2. \(y_2 - y_1 = 1 - (-7) = 8\) Now, substituting back into the formula: \[ d = \sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} \] Rounding to the tenths place, we find: \[ d \approx 8.2 \] Next, let's find the distance between the points \((-6, -10)\) and \((-2, -10)\): Using the distance formula again: 1. \(x_2 - x_1 = -2 - (-6) = 4\) 2. \(y_2 - y_1 = -10 - (-10) = 0\) Now calculate: \[ d = \sqrt{4^2 + 0^2} = \sqrt{16} = 4 \] Thus, the distances rounded to the tenths place are approximately: 1. **For points \((-10, -7)\) and \((-8, 1)\)**: 8.2 2. **For points \((-6, -10)\) and \((-2, -10)\)**: 4.0