What is the distance between the points (2,3) and (-1,0)?

Answer 1

Distance formula

d = √(x₂- x₁)² + (y₂- y₁)²

Ordered pairs

(2, 3)(-1, 0)

d = √(-1 -2)² + (0 - 3)²

d = √(-3)² + (3)²

d = √9 + 9

d = √18

d = √9 √2

d = 3 √2

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Answer 2

The formula to find the distance between two points (x1, y1), (x2, y2) is
D=√(x2-x1)²+ (y2-y1)² therefore the answer for your question will be
(2, 3), (-1, 0)..._
D= √ (-1-2)² + (0-3)²
D=√18
D=3 √2
It is also the same as finding the hypotenuse of a right angle triangle with legs measuring 3 units each ..

Answer 3

3√2 units

So you can visualize the problem (not try to memorize), draw your standard x,y graph, plot your points, subtract x2-x1, y2-y1. On your graph you can clearly see that the distance is the hypotenuse of a triangle with side x (-1 to 2) being 3 units long, and side y (0 to 3) also being 3 units long.

Using Pythagorus's Theorem,
d = √(x² + y²) = √(9 + 9) = √18 = √(9)(2) = 3√2.*******

***Judah below is wrong...he's trying to plug into a formula instead of plotting the problem on a graph so he can see it--and he's got the formula wrong...he forgot to take the square root.***

Answer 4

The formula for distance goes as follows

d = sqrt ( (x2 - x1)^2 + (y2 - y1)^2 )

In our case, (x1, y1) = (2, 3) and (x2, y2) = (-1, 0), so we have

d = sqrt ( (-1 - 2)^2 + (0 - 3)^2 )
d = sqrt( (-3)^2 + (-3)^2 )
d = sqrt(9 + 9)
d = sqrt(18)

This can be reduced by radical-reducing methods as normal.

sqrt(18) = sqrt(9 * 2) = sqrt(9)sqrt(2) = 3sqrt(2)

Therefore

d = 3sqrt(2)

Answer 5

plot these points on a graph.
so you need to find the distance between the x-components and the distance between the y-components.
x-components: 2-(-1)=3
y-components: 3-0=3
With pythagoras, sqrt(3^2 + 3^2) = 3*sqrt(2)

Answer 6

18. i used the distance formula
(x1 - x2) squared + (y1 - y2) squared

Answer 7

SQRT 18

Answer 8

(3,3)