Find the distance d(P1,P2)between the points P1 and P2.
P1=(-5,2)
P2=(2,1)
d(P1,P2)=?
*all the 1's and 2's are to be lower down.* Please show work
The distance between the X coordinates is 2 minus -5 = 7. The distance between the Y coordinates is 2 minus 1 = 1. Those two distances are the base and height of a right-angled triangle, so to get the hypoteneuse, use Pythagoras: that gives you that the distance is the square root of (7*7 + 1*1) = sqrt(50) = roughly 7.071. I don't know where the calculus comes in, though.
The distance between the P1 (8, 4) and P2 (5, 2) is 5.
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2 points
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To find the distance between two points, we can use the distance formula:
d(P1, P2) = √((x2 - x1)^2 + (y2 - y1)^2)
Given that P1 = (-5,2) and P2 = (2,1), we can substitute the coordinates into the formula:
d(P1, P2) = √((2 - (-5))^2 + (1 - 2)^2)
simplifying:
d(P1, P2) = √((2 + 5)^2 + (-1)^2)
d(P1, P2) = √(7^2 + 1)
d(P1, P2) = √(49 + 1)
d(P1, P2) = √50
The distance between P1 and P2 is √50.
To find the distance between two points in a coordinate plane, we can use the distance formula:
d(P1, P2) = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, P1 has coordinates (-5,2) and P2 has coordinates (2,1). Let's substitute those values into the distance formula:
d(P1, P2) = √((2 - (-5))^2 + (1 - 2)^2)
Now, let's simplify:
d(P1, P2) = √((2 + 5)^2 + (-1)^2)
d(P1, P2) = √(7^2 + 1^2)
d(P1, P2) = √(49 + 1)
d(P1, P2) = √50
Finally, we can simplify the square root:
d(P1, P2) = √(25 * 2)
d(P1, P2) = √25 * √2
d(P1, P2) = 5 * √2
Therefore, the distance d(P1, P2) between points P1 and P2 is 5√2 (approximately 7.071).