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vazorg [7]
3 years ago
7

Locate the point of the line segment between A (3, -5) and B (13, -15) given that the point is 4/5 of the way from A to B. Show

your work.

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
3 0

Answer:

(11, -13)

Step-by-step explanation:

A (3, -5)  B (13, -15)

AC = |difference of y = |(-15 - (-5)| = 10

CB = | difference of x| = |13 - 3| = 10

ΔAMN ~ ΔACB ---- AA

∴ AM / AC = MN / CB = AN / AB

AM / 10 = MN / 10 = (4/5 AB) / AB = 4/5

AM = MN = 8

coordinate of M:    y coordinate = -13 ... same as y coordinate of N

coordinate of N: x coordinate = 3 + 8 = 11

The point 4/5 the way from A to B is (11 , -13)

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Write the equation for the line parallel to the given line 4x-3y=9 and through the point (3, -1). Then Write the answers in slop
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k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ \perp\ l\iff m_1\cdot m_2=-1\\\\k\ ||\ l\iff m_1=m_2

We have:

4x-3y=9\ \ \ \ |-4x\\\\-3y=-4x+9\ \ \ \ |:(-3)\\\\y=\dfrac{4}{3}x-3\to m_1=\dfrac{4}{3}

y=m_2x+b\\\\m_2=m_1\to m_2=\dfrac{4}{3}

y=\dfrac{4}{3}x+b

The line passest through the point (3, -1). Substitute the coordinates of the point to the equation of a line:

-1=\dfrac{4}{3}\cdot3+b\\\\-1=4+b\ \ \ \ |-4\\\\b=-5

y=\dfrac{4}{3}x-5\ \ \ \ |\cdot3\\\\3y=4x-15\ \ \ \ |-4x\\\\-4x+3y=-15\ \ \ \ |\cdot(-1)\\\\4x-3y=15

Answer: \boxed{y=\dfrac{4}{3}x-5;\ 4x-3y=15}

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