Question

find the distance between the two points in simplest radical form.

Explanation:

Step1: Identify coordinates

First, find the coordinates of the two points. The upper point is at \((-7, 5)\) and the lower point is at \((2, -7)\).

Step2: Apply distance formula

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Substitute \(x_1 = -7\), \(y_1 = 5\), \(x_2 = 2\), \(y_2 = -7\):

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Wait, no, wait. Wait, let's recalculate the differences. \(x_2 - x_1 = 2 - (-7) = 9\), \(y_2 - y_1 = -7 - 5 = -12\). Then squares: \(9^2 = 81\), \((-12)^2 = 144\). Sum: \(81 + 144 = 225\). Square root of 225 is 15. Wait, but maybe I misread the coordinates. Wait, the lower point: looking at the graph, the x-coordinate is 2, y-coordinate is -7? Wait, no, let's check again. The lower point: x is 2, y is -7? Wait, the grid: from the origin (0,0), moving right 2 units (x=2), down 7 units (y=-7). The upper point: x=-7 (left 7 units), y=5 (up 5 units). So the differences: \(\Delta x = 2 - (-7) = 9\), \(\Delta y = -7 - 5 = -12\). Then distance is \(\sqrt{9^2 + (-12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15\). Wait, but that seems straightforward. Wait, maybe I made a mistake in coordinates? Let me check again. The upper point: x=-7, y=5 (since x=-7 is 7 units left of origin, y=5 is 5 units up). The lower point: x=2 (2 units right), y=-7 (7 units down). So yes, \(\Delta x = 2 - (-7) = 9\), \(\Delta y = -7 - 5 = -12\). Then distance is \(\sqrt{9^2 + 12^2}\) (since square of -12 is 144, same as 12^2). So \(\sqrt{81 + 144} = \sqrt{225} = 15\). Wait, but 9-12-15 is a Pythagorean triple (3-4-5 scaled by 3: 9=33, 12=34, 15=3*5). So that makes sense.

Answer:

\(15\)