Explanation:
these are triangles.
for triangle EFG we have
EF = sqrt((6 - 9)² + (2 - 2)²) = sqrt(9) = 3
EG = sqrt((6 - 8)² + (2 - 7)²) = sqrt(4 + 25) = sqrt(29)
FG = sqrt((9 - 8)² + (2 - 7)²) = sqrt(1 + 25) = sqrt(26)
for triangle HIJ we have
HI = sqrt((0 - 0)² + (0 - 3)²) = sqrt(9) = 3
HI corresponds to EF.
now, J = (x, y) with the same side lengths to H and I, as G to E and F.
so,
HJ = sqrt((0 - x)² + (0 - y)²) = sqrt(29)
but it could be also
HJ = sqrt((0 - x)² + (0 - y)²) = sqrt(26)
in the same way
IJ = sqrt((0 - x)² + (3 - y)²) = sqrt(29)
but it could be also
IJ = sqrt((0 - x)² + (3 - y)²) = sqrt(26)
we just have to make sure that they are different.
for HJ we get
sqrt(x² + y²) = sqrt(29) or sqrt(26)
x² + y² = 29 or 26
for IJ we get
sqrt(x² + (3 - y)²) = sqrt(29) or sqrt(26)
x² + (3 - y)² = 29 or 26
x² + 9 - 6y + y² = 29 or 26
let's start with
x² + y² = 29
y² = 29 - x²
y = sqrt(29 - x²)
this gives us for the second case
x² + 9 - 6y + y² = 26
x² + 9 - 6sqrt(29 - x²) + 29 - x² = 26
-6sqrt(29 - x²) + 29 = 17
-6sqrt(29 - x²) = -12
sqrt(29 - x²) = 2
29 - x² = 4
-x² = -25
x² = 25
x = ±5
remember, the solution to a square root always has a positive and a negative value, as the square of them is the same.
out of
x² + y² = 29
we get now
5² + y² = 29
25 + y² = 29
y² = 4
y = ±2
out of the ±5, ±2 combinations we need to verify also with
x² + 9 - 6y + y² = 26
we see, x = ±5, y = +2 this is correct
25 + 9 - 12 + 4 = 26
but for x = ±5, y = -2 this fails
25 + 9 + 12 + 4 = 26
50 = 26 is wrong.
so, we get for
HJ = sqrt(29), IJ = sqrt(26) the possible solutions
J = (5, 2), J = (-5, 2)
now, let's look at
x² + y² = 26
y² = 26 - x²
y = sqrt(26 - x²)
this gives us for the second case
x² + 9 - 6y + y² = 29
x² + 9 - 6sqrt(26 - x²) + 26 - x² = 29
-6sqrt(26 - x²) + 26 = 20
-6sqrt(26 - x²) = -6
sqrt(26 - x²) = 1
26 - x² = 1
-x² = -25
x² = 25
x = ±5
remember, the solution to a square root always has a positive and a negative value, as the square of them is the same.
out of
x² + y² = 26
we get now
5² + y² = 26
25 + y² = 26
y² = 1
y = ±1
out of the ±5, ±1 combinations we need to verify also with
x² + 9 - 6y + y² = 29
we see, x = ±5, y = +1 this is correct
25 + 9 - 6 + 1 = 29
but for x = ±5, y = -1 this fails
25 + 9 + 6 + 1 = 29
41 = 29 is wrong.
so, we get for
HJ = sqrt(26), IJ = sqrt(29) the possible solutions
J = (5, 1), J = (-5, 1)
we can say the 4 possible solutions are caused by the same solution for J, one on the left, and one on the right side of HI. and one for HJ = sqrt(29), and one for HJ = sqrt (26), as nobody defined the correlation of the legs. it could be up or down.
so, in total 2×2 = 4 solutions.