To the nearest hundredth, what is the distance between point (-6, -1) and point (4, 3)?
units
Calculating the distance between two points in a coordinate plane: To the nearest hundredth, what is the distance between point (-6, -1) and point (4, 3)?
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To find the distance between the two points (-6, -1) and (4, 3), we can use the distance formula:
[
d = sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}
]
Here, ((x_1, y_1) = (-6, -1)) and ((x_2, y_2) = (4, 3)).
1. Calculate (x_2 – x_1) and (y_2 – y_1):
[
x_2 – x_1 = 4 – (-6) = 4 + 6 = 10
]
[
y_2 – y_1 = 3 – (-1) = 3 + 1 = 4
]
2. Plug these values into the distance formula:
[
d = sqrt{(10)^2 + (4)^2}
]
[
d = sqrt{100 + 16} = sqrt{116}
]
3. Calculate (sqrt{116}):
[
d approx 10.77 text{ (to the nearest hundredth)}
]
Therefore, the distance between the points is approximately 10.77 units.
To find the distance between two points ((-6, -1)) and ( (4, 3)), we can use the distance formula:
[
d = sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}
]
Here, ((x_1, y_1) = (-6, -1)) and ((x_2, y_2) = (4, 3)).
1. Calculate (x_2 – x_1):
[
4 – (-6) = 4 + 6 = 10
]
2. Calculate (y_2 – y_1):
[
3 – (-1) = 3 + 1 = 4
]
3. Substitute into the formula:
[
d = sqrt{(10)^2 + (4)^2} = sqrt{100 + 16} = sqrt{116}
]
4. Calculate (sqrt{116}):
[
sqrt{116} approx 10.77
]
Therefore, the distance between the points ((-6, -1)) and ( (4, 3)) is approximately 10.77 units.