(x+4)^2 + (y-3)^2 = 29
So 29 has to be sum of two squares.
29 = 25 + 4
Those are the only two integers.
so ..
Case 1: (x+4)^2 = 25 and (y-3)^2 = 4
or
Case 2: (x+4)^2 = 4 and (y-3)^2 = 25
Case 1:
then x + 4 = ±2 and y-3 =±5
x = -2, y = 8
x = -6, y = 8
x = -2 , y = -2
x = -6 , y = -2
Case 1:
x+4 =±5 and y-3 = ±2
x = -1 , y = 5
x = -1 , y = 1
x = -9 , y = 5
x = -9 , y = 1
Case 2:
then x + 4 = ±2 and y-3 =±5
x = -2, y = 8
x = -6, y = 8
x = -2 , y = -2
x = -6 , y = -2
So it looks like 8 points.
Find all points with integer coordinates that are a distance of square root of 29 from (-4,3)
how do u do this?????!?!?!?!
2 answers
(x+4)^2 + (y-3)^2 = 29
So 29 has to be sum of two squares.
29 = 25 + 4
Those are the only two integers.
so ..
Case 1: (x+4)^2 = 25 and (y-3)^2 = 4
or
Case 2: (x+4)^2 = 4 and (y-3)^2 = 25
Case 1:
x+4 =±5 and y-3 = ±2
x = 1 , y = 5
x = 1 , y = 1
x = -9 , y = 5
x = -9 , y = 1
Case 2:
then x + 4 = ±2 and y-3 =±5
x = -2, y = 8
x = -6, y = 8
x = -2 , y = -2
x = -6 , y = -2
So it looks like 8 points.
So 29 has to be sum of two squares.
29 = 25 + 4
Those are the only two integers.
so ..
Case 1: (x+4)^2 = 25 and (y-3)^2 = 4
or
Case 2: (x+4)^2 = 4 and (y-3)^2 = 25
Case 1:
x+4 =±5 and y-3 = ±2
x = 1 , y = 5
x = 1 , y = 1
x = -9 , y = 5
x = -9 , y = 1
Case 2:
then x + 4 = ±2 and y-3 =±5
x = -2, y = 8
x = -6, y = 8
x = -2 , y = -2
x = -6 , y = -2
So it looks like 8 points.