The point (1,6) is one endpoint of a line segement that has a length
of square root of 80 units. The other endpoint is in quadrant III and
has an x-coordinate of -3.
We can draw a right triangle using these two points, and use the
Pythagorean theorem to find what we are looking for.
The length of the x leg is (1 - -3), which is 4.
The length of the y leg is (6 - y).
We are given the length of the hypotenuse, the square root of 80.
The formula is x^2 + y^2 = c^2 (I used x and y in place of a and b)
4^2 + (6 - y)^2 = 80 (square root of 80, squared)
16 + 36 - 12y + y^2 = 80
Subtract 80 from both sides.
16 + 36 - 12y + y^2 - 80 = 80 - 80
y^2 - 12y - 80 + 36 + 16 = 0
y^2 - 12y - 28 = 0
Factoring, we get the following:
(y - 14) * (y + 2) = 0
Because we are multiplying and getting a zero, one of
the two factors has to be zero.
y - 14 = 0
y = 14
Or...
y + 2 = 0
y = -2
The problem statement says the endpoint we are interested in
is in quadrant III, which means the y co-ordinate has to be
negative, therefore y = -2
The point is (-3, -2)