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A line sgement has endpoints at (-3,2) amd (42,32). What is the x-coordinate of the point that is 2/3 the distance from (-3, 2) to (42, 32)

User Run
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Answer:x = midpoint_x - (midpoint_x - (-3)) * (sqrt(5)/6) = 19.5 - (22.5 * sqrt(5))/6 = 19.5 - 3.75sqrt(5)

Step-by-step explanation:To find the x-coordinate of the point that is 2/3 of the distance from (-3, 2) to (42, 32), we first need to find the coordinates of that point.

The distance between the two endpoints of the line segment is:

d = sqrt((42 - (-3))^2 + (32 - 2)^2) = sqrt(45^2 + 30^2) = 15sqrt(45)

The distance from (-3, 2) to the desired point is:

(2/3)d = (2/3)sqrt(45^2 + 30^2)

Let's call the desired point (x, y). We can use the midpoint formula to find the coordinates of the midpoint of the line segment:

midpoint = ((-3 + 42)/2, (2 + 32)/2) = (19.5, 17)

The midpoint is halfway between the two endpoints, so the distance from (-3, 2) to the midpoint is:

sqrt((19.5 - (-3))^2 + (17 - 2)^2) = sqrt(22.5^2 + 15^2) = 15sqrt(5)

To find the coordinates of the desired point, we can use similar triangles. The ratio of the distance from (-3, 2) to the midpoint to the distance from (-3, 2) to the desired point is:

15sqrt(5) / (2/3)sqrt(45^2 + 30^2) = 15sqrt(5) / 30sqrt(45) = sqrt(5)/6

The ratio of the distance from (42, 32) to the midpoint to the distance from (42, 32) to the desired point is the same:

15sqrt(5) / (2/3)sqrt(45^2 + 30^2) = sqrt(5)/6

User Grozz
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