asked 76.4k views
2 votes
AB has endpoints A (-3Ė9) and B (7Ė6) . Point C divides

AB into two parts with the lengths in a ratio of 3:2. Wha
tis the location of C?
=
numerator
numerator+deno min ator
P (xĖy) = (x1+k(x2 - X1), Y1 + k(y2 -y1)

1 Answer

2 votes

Point C divides line segment AB into two parts with lengths in a ratio of 3:2. Let's first calculate the length of AB. Using the distance formula, we get:

AB = √[(7 - (-3))^2 + (6 - 9)^2] = √(10^2 + 3^2) = √109

Let's assume that C divides AB into two parts with lengths 3k and 2k, respectively. Then, we have:

AC = 3k and BC = 2k

Since C is on the line segment AB, we can write:

AC + BC = AB

3k + 2k = √109

5k = √109

k = √109 / 5

Now, we can find the coordinates of point C using the midpoint formula:

C(x, y) = [(3k * x2 + 2k * x1) / (3k + 2k), (3k * y2 + 2k * y1) / (3k + 2k)]

Substituting the given values, we get:

C(x, y) = [(3/5) * 7 + (2/5) * (-3), (3/5) * 6 + (2/5) * 9] = [1.4, 6.6]

Therefore, point C is located at **(1.4, 6.6)**.

answered
User Bitcycle
by
7.2k points
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