SOLUTION: find y if the distance between points p and R is 25 and point R is located in the first quadrant. P=(3,-8) R=(10,y)

Algebra ->  Coordinate-system -> SOLUTION: find y if the distance between points p and R is 25 and point R is located in the first quadrant. P=(3,-8) R=(10,y)      Log On


   



Question 1139586: find y if the distance between points p and R is 25 and point R is located in the first quadrant.
P=(3,-8)
R=(10,y)

Found 4 solutions by ikleyn, MathLover1, Theo, greenestamps:
Answer by ikleyn(52776) About Me  (Show Source):
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.
%2810-3%29%5E2 + %28y-%28-8%29%29%5E2 = 25%5E2


7%5E2 + %28y%2B8%29%5E2 = 625


49 + y%2B8%29%5E2 = 625


%28y%2B8%29%5E2 = 625 - 49 = 576


y + 8 = +/- sqrt%28576%29 = +/- 24


y = -8 +/- 24.


y%5B1%5D = -8 + 24 = 16;   y%5B2%5D = -8 - 24 = -32.


Only positive  y%5B1%5D = 16 provides R is located in QI.


ANSWER.  y = 16.

Solved.


Answer by MathLover1(20849) About Me  (Show Source):
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use distance formula:
d=sqrt%28%28x-x%5B1%5D%29%5E2%2B%28y-y%5B1%5D%29%5E2%29
given:
P=(3,-8)=(x%5B1%5D,y%5B1%5D)
R=(10,y) =(x,y)
the distance between points P and R is d=25
25=sqrt%28%2810-3%29%5E2%2B%28y-%28-8%29%29%5E2%29
25=sqrt%28%287%29%5E2%2B%28y%2B8%29%5E2%29
25=sqrt%2849%2B%28y%2B8%29%5E2%29.......square both sides
25%5E2=49%2B%28y%2B8%29%5E2......solve for y
625-49=%28y%2B8%29%5E2
576=%28y%2B8%29%5E2
sqrt%28576%29=y%2B8
24-8=y
y=16
=> answer: R=(10,16) which is located in the first quadrant


Answer by Theo(13342) About Me  (Show Source):
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point p is at (3,-8)
point r is at (10,y)

the distance between them is 25.

point r is in the first quadrant.

this means that y has to be positive.

the distance between point p and point r is equal to sqrt((y+8)^2 + (10-3)^2).

simplify this to get distance between points p and r is equal to sqrt((y+8)^2 + 49)

since the distance between points p and r is 25, then the formula becomes:

25 = sqrt((y+8)^2 + 49)

square both sides of the equation to get 625 = (y+8)^2 + 49

simplify to get 625 = y^2 + 16y + 64 + 49

combine like terms to get 625 = y^2 + 16y + 113

subtract 625 from both sides of the equation to get 0 = y^2 + 16y - 512.

factor this quadratic equation to get (y + 32) * (y - 16) = 0

solve for y to get y = -32 or 16.

y is positive, so y has to be 16.

your solution is that y = 16.

this means that point p = (3,-8) and point r = (10,16)

the distance between points p and r is equal to sqrt((16+8)^2 + (10-3)^2).

that becomes equal to sqrt((24)^2 + 7^2) which becomes equal to sqrt(625) which becomes equal to 25.

that confirms that, when y = 16, the distance between p and r is 25.

the equation of the line between points p and r is y = 24/7 * x -128/7.

the graph of that equation is shown below.

it shows that the points (3,-8) and (10,16) are both on the line, as they sh ould be.

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Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The distance between the two points is 25; the difference between the x-coordinates is 7.

7-24-25 is a Pythagorean triple, so the difference between the y-coordinates must be 24.

If R is to be in the first quadrant, y must be -8+24 = 16.

ANSWER: y = 16