Question

reminder: be sure to simplify the radical to get your final answer.
what is the distance between points v and w?
v(-1,5)
w(-6,-4)
options: (2sqrt{65}), (2sqrt{53}), (sqrt{130}), (sqrt{106})

Explanation:

Step1: Recall Distance Formula

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).

Step2: Identify Coordinates

Point \(V\) is \((-1, 5)\) so \(x_1 = -1\), \(y_1 = 5\). Point \(W\) is \((-6, -4)\) so \(x_2 = -6\), \(y_2 = -4\).

Step3: Substitute into Formula

Calculate \(x_2 - x_1 = -6 - (-1) = -5\) and \(y_2 - y_1 = -4 - 5 = -9\). Then \(d = \sqrt{(-5)^2 + (-9)^2} = \sqrt{25 + 81} = \sqrt{106}\)? Wait, no, wait: Wait, \((-5)^2 = 25\), \((-9)^2 = 81\), sum is \(106\)? Wait, no, wait the options have \(2\sqrt{65}\), let's recalculate. Wait, maybe I mixed up points. Wait \(V(-1,5)\), \(W(-6,-4)\). So \(x_1=-1\), \(x_2=-6\): difference is \(-6 - (-1) = -5\), absolute value squared is 25. \(y_1=5\), \(y_2=-4\): difference is \(-4 -5 = -9\), squared is 81. Sum is 25 + 81 = 106? But the options: wait, maybe I made a mistake. Wait, no, let's check again. Wait, maybe the points are \(V(-1,5)\) and \(W(-6,-4)\). Wait, distance formula: \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). So \(x_2 - x_1 = -6 - (-1) = -5\), \(y_2 - y_1 = -4 - 5 = -9\). Then squared terms: 25 and 81. Sum is 106. But the options have \(2\sqrt{65}\), \(\sqrt{130}\), etc. Wait, maybe I swapped the points? Wait, no, let's check the graph. Wait, maybe \(V\) is \((-1,5)\) and \(W\) is \((-6,-4)\). Wait, no, maybe I miscalculated. Wait, \((-1 - (-6)) = 5\), and \(5 - (-4) = 9\). Oh! Wait, I had \(x_1\) and \(x_2\) reversed. Let's correct: \(x_1 = -6\), \(y_1 = -4\) (point \(W\)), \(x_2 = -1\), \(y_2 = 5\) (point \(V\)). Then \(x_2 - x_1 = -1 - (-6) = 5\), \(y_2 - y_1 = 5 - (-4) = 9\). Then distance is \(\sqrt{5^2 + 9^2} = \sqrt{25 + 81} = \sqrt{106}\)? No, that's the same. Wait, the options: let's check the options again. The options are \(2\sqrt{65}\), \(2\sqrt{53}\), \(\sqrt{130}\), \(\sqrt{106}\). Wait, maybe I made a mistake in coordinates. Wait, looking at the graph: \(V\) is at \((-1,5)\), \(W\) is at \((-6,-4)\). So horizontal distance: from \(x=-1\) to \(x=-6\) is 5 units (since \(-1 - (-6) = 5\)). Vertical distance: from \(y=5\) to \(y=-4\) is 9 units (5 - (-4) = 9). Then distance is \(\sqrt{5^2 + 9^2} = \sqrt{25 + 81} = \sqrt{106}\). But the options have \(\sqrt{106}\) as one of them? Wait, the last option is \(\sqrt{106}\)? Wait, the options are: first \(2\sqrt{65}\), second \(2\sqrt{53}\), third \(\sqrt{130}\), fourth \(\sqrt{106}\). So the correct answer should be \(\sqrt{106}\)? Wait, but let's recalculate. Wait, 5 squared is 25, 9 squared is 81. 25 + 81 is 106. So \(\sqrt{106}\) is the distance. So the correct option is the last one, \(\sqrt{106}\).

Answer:

\(\sqrt{106}\) (the last option, e.g., if options are labeled as A: \(2\sqrt{65}\), B: \(2\sqrt{53}\), C: \(\sqrt{130}\), D: \(\sqrt{106}\), then D. \(\sqrt{106}\))