To avoid doing people's homework, I'll work a problem exactly like yours,
but with different numbers.
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Find the coordinates of P if P is 17 units from L(-3,-6) and 10 units from M(-1,1)
It appears that there are two solutions.
OR
We use the distance formula, setting PL = 17 and PM = 10
d = = 17
= 10
Squaring both sides of both equations:
(x+3)² + (y+6)² = 17²
(x+1)² + (y-1)² = 10²
x²+6x+9 + y²+12y+36 = 289
x²+2x+1 + y²-2y+1 = 100
Simplifying
x²+y²+6x+12y = 244
x²+y²+2x- 2y = 98
Subtracting those two equations:
4x+14y = 146
Divide that equation through by 2
2x+ 7y = 73
2x = 73-7y
x =
Substitute in
x²+y²+2x- 2y = 98
+ y² + 2 - 2y = 98
+ y² + 73 - 7y - 2y = 98
+ y² - 9y = 25
Clear the fraction by multiplying through by 4
5329 - 1022y + 49y² + 4y² - 36y = 100
53y² - 1058y + 5229 = 0
That factors as
(53y - 581)(y - 9) = 0
53y - 581 = 0; y - 9 = 0
53y = 581; y = 9
y =
Substitute each in 2x + 7y = 73
2x + 7y = 73
2x + 7() = 73
106x + 4067 = 3869
106x = -198
x =
x =
So one solution is the point P(, )
[That's approximately the point P(-1.97,10.96)
The other solution is
2x + 7y = 73
2x + 7(9) = 73
2x + 63 = 73
2x = 10
x = 5
So the other solution is P(5,9)
Edwin