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what is the distance between the points shown on the coordinate plane b…

Question

what is the distance between the points shown on the coordinate plane below? provide an answer accurate to the nearest tenth.

Explanation:

Step1: Identify coordinates

First, find the coordinates of the two points. The upper point is at \((-1, 3)\) (wait, looking at the grid: the vertical line is y-axis, horizontal x-axis. The upper dot: x is -1 (since it's 1 unit left of y-axis), y is 3 (3 units above x-axis). The lower dot is at \((0, -1)\) (on y-axis, x=0, y=-1). Wait, maybe better to check again. Wait, the grid: each square is 1 unit. Let's re-express:

Wait, the upper point: x-coordinate: let's see, from y-axis (x=0), moving left 1 unit: x=-1. y-coordinate: from x-axis (y=0), moving up 3 units: y=3. So \((-1, 3)\).

The lower point: on y-axis, so x=0, y=-1 (since it's 1 unit below x-axis). So \((0, -1)\).

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 \(x_1 = -1\), \(y_1 = 3\), \(x_2 = 0\), \(y_2 = -1\):

\(d = \sqrt{(0 - (-1))^2 + (-1 - 3)^2}\)

Simplify inside the square root:

\(0 - (-1) = 1\), \(-1 - 3 = -4\)

So \(d = \sqrt{(1)^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17}\)

Step3: Calculate and round

\(\sqrt{17} \approx 4.123\), round to nearest tenth: 4.1

Wait, wait, maybe I misread the coordinates. Let me check again. Wait, the upper point: looking at the grid, the x-axis: from -6 to 6, y-axis from -6 to 6. The upper dot: x is -1? Wait, no, maybe the upper point is at (-1, 3)? Wait, no, maybe the upper point is at (-1, 3) and the lower at (0, -1). Wait, or maybe the upper point is at (-1, 3) and the lower at (0, -1). Wait, let's re-express:

Wait, the lower point is at (0, -1) (on y-axis, x=0, y=-1). The upper point: x is -1 (1 unit left of y-axis), y is 3 (3 units above x-axis). So the horizontal distance (Δx) is \(0 - (-1) = 1\), vertical distance (Δy) is \(-1 - 3 = -4\), so squared distances: \(1^2 = 1\), \((-4)^2 = 16\), sum is 17, square root of 17 is approx 4.123, so to nearest tenth is 4.1.

Wait, but maybe the coordinates are different. Wait, maybe the upper point is at (-1, 3) and the lower at (0, -1). Let me confirm:

Looking at the grid: the upper dot is at x=-1 (since it's one square left of y-axis), y=3 (three squares above x-axis). The lower dot is at x=0 (on y-axis), y=-1 (one square below x-axis). So yes, coordinates are (-1, 3) and (0, -1).

So distance is \(\sqrt{(0 - (-1))^2 + (-1 - 3)^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.1\)

Answer:

4.1