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Question 1118318: Find all the points having an x-coordinate of 9 whose distance from the point (3, -2) is 10.
(9, 13), (9, -7)
(9, 2), (9, -4)
(9, 6), (9, -10)
(9, -12), (9, 8)
Found 2 solutions by htmentor, greenestamps:
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! The distance between two points is given by:
d = sqrt((x1-x2)^2+ (y1-y2)^2)
Given: x-coordinate is equal to 9, and the distance is equal to 10
Thus we have
10^2 = (9-3)^2 + (y+2)^2
(y+2)^2 = 64
Solve for y:
y^2 + 4y - 60 = 0
Factors as (y+10)(y-6) = 0
Solutions are y = 6, y = -10
Ans: (9,6), (9,-10)
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
The formal algebraic solution method which has you finding the answer by solving a quadratic equation is of course valid. But the problem can be made much easier than that.
You are looking for the two points on the line x=9 whose distance from (3,-2) is 10.
The horizontal distance from (3,-2) to the line x=9 is 9-3 = 6. So that horizontal line segment, the line x=9, and the segments from (3,-2) to the points we are looking for will form two right triangles, each with one leg 6 and hypotenuse 10.
Quick use of the Pythagorean Theorem (or the recognition that 6-8-10 is a multiple of the common 3-4-5 right triangle) make the other leg of each right triangle 8.
Then the two points on the line x=9 that are 8 units from y=-2 are -2+8 = 6 and -2 -8 = -10.
So the two points we are looking for are (9,-10) and (9,6).
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