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Eva8 [605]
2 years ago
7

Point A is at ( -3,4) and Point C is at ( 2,-6 )

Mathematics
1 answer:
marishachu [46]2 years ago
4 0

The required coordinates of point B will be (2/9, -14/9).

Point A is at ( -3,4) and Point C is at ( 2,-6 )

Given that point B on line, AC suck as the ratio of AB to AC is 4:5

The coordinates of point B will be,

{[(mx2+nx1)/(m+n)],[(my2+ny1)/(m+n)]}

Breaking it down, the x coordinate is (mx2+nx1)/(m+n) and the y coordinate is (my2+ny1)/(m+n)

As per the question, m = 4 and n = 5

The coordinates of point B :

⇒ {[(4×-3+5×2)/(4+5)],[(4×4+5×-6)/(4+5)]}

⇒ {[(-12+10)/(9)],[(16-30)/(9)]}

⇒ (2/9, -14/9)

Thus, the required coordinates of point B will be (2/9, -14/9).

Learn more about the distance between two points here:

brainly.com/question/15958176

#SPJ1

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One side of a triangle is 6 cm shorter than the base, x. The other side is 4 cm longer than the base. What lengths of the base w
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X= base
one side= x-6
other side= 4+x

x+x-6+4+x = 43

3x=43+2

x=15 length of base
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3 years ago
Find the unit vectors that are parallel to the tangent line to the curve y = 8 sin(x) at the point (Ï/6, 4). (enter your answer
Airida [17]
Given:
y = 8 sin(x)

y'(x) = 8 cos(x)
At the point(1/6, 4), obtain the slope as
y' = 8 cos(1/6) = 7.889

Parallel line:
A line parallel to the curve at (1/6, 4) will have the same slope.
Let the line be
y = 7.889x + c
Because the line passes through(1/6, 4), therefore
7.889(1/6) + c = 4   =>  c = 2.6852
The line is y = 7.889x + 2.6852
Two points on the line are (0, 2.6852) and (1, 10.5742)
The distance between the points is √(1 + 7.889²) = 7.952
The umit vector is (0.126, 0.992)

Perpendicular line:
The product of the slopes of the parallel and perpendicular line is -1.
Therefore the slope of the perpendicular line is - 1/7.889 = - 0.1268.
Let the perpendicular line be
y = -0.1268x + c
Because the line passes through (1/6, 4), therefore
-0.1268(1/6) + c = 4  =>  c = 4.0211
The line is y = -0.1268x + 4.0211
Two points on the line are (0, 4.0211) and (1, 3.8943).
The distance between the points is √(1 + (-0.1268)²) = 1.008
The unit vector is (0.992, -0.126)

A graph of the curve and the two lines is shown below.

Answer:
The unit vector for the parallel line is (0.126, 0.992)
The unit vector for the perpendicular line is (0.992, -0.126)

4 0
3 years ago
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