| Post
Answer |
1. Given A(6, -4), find the coordinates of the point B, such that C(3, -2) is the midpoint of segment AB. Need answer as an ordered pair in the form (x,y).
2.Find a formula that expresses the fact that an arbitrary point P(x, y) is on the perpendicular bisector l of segment AB. Need answer in the form ax + by = c, where a, b and c are integers. Make sure that you have divided out any common factors of a, b and c.
3.Find a formula that expresses the fact that P(x, y) is a distance 3 from the origin.
4.Find the point with coordinates of the form (2a, a) that is in the third quadrant and is a distance 10 from P(2, 6).
I would greatly appreciate this. Thank you in advance! |
| Answer |
1) Start at (6, -4). To get to (3, -2) from there, you need to go left 3 units and up 2. Since (3, -2) is the midpoint, to find point B you can just repeat this movement: Start at (3, -2) and go left 3 units and up 2. You will be at (0,0), the origin.
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2) I\'m not quite sure what they want here. However, if you draw a line segment, AB, and then draw its perpendicular bisector, you will notice that any point (x,y) on the bisector must be equidistant from point A and point B. Since the other problems you have listed here involve finding the distance between points, I believe this is relevant to what you are trying to do. Maybe you could call point A, (xa,ya) and point B, (xb,yb). Then [(x -xa)^2 + (y - ya)^2]^(1/2) = [(x - xb)^2 + (y - yb)^2]^(1/2). You may be able to shuffle that information around in order to arrive at an equation. I\'d have to look at it some more to see if that is the way to go.
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3) P(x,y) is a distance of 3 from the origin means:
[(x - 0)^2 + (y - 0)^2]^(1/2) = 3
--> (x^2 + y^2)^(1/2) = 3 -->
x^2 + y^2 = 3^2 -->
x^2 + y^2 = 9 {You will notice that this is the equation of a circle of radius = 3 centered at (0,0). }
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4) Point (2a,a) is a distance 10 from P(2,6) means:
[(2 - 2a)^2 + (6 - a)^2]^(1/2) = 10
Expand:
--> [(4 - 8a + 4a^2) + (36 - 12a + a^2)]^(1/2) = 10
Combine terms on the left side of the equation:
[5a^2 - 20a + 40]^(1/2) = 10
Square both sides:
5a^2 - 20a + 40 = 100
Divide by 5:
a^2 - 4a + 8 = 20
Subtract 20 from both sides:
a^2 - 4a - 16 = 0 Solve for a using the quadratic formula ( x = [ -b +/- (b^2 - 4ac)^(1/2)]/2a )
a=(4+/-[(-4)^2 - 4(1)(-16)]^.5)/2(1) -->
a = (4 +/- (80)^.5)/2 -->
a = (4 +/- 4(5^.5))/2 -->
a = 2 +/- 2 (5^.5)
Since the problem states that (2a, a) is in the 3rd quadrant, this means both 2a and a must be negative. Since 2 + 2(5^.5) is positive, this is not the answer. Therefore, a must be 2 - 2(5^.5). The point you are looking for will be: (x,y) = (4 - 4(5^.5), 2 - 2(5^.5))
I hope this was helpful! |
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