Question

Find all y such that the distance between the points (1, 10) and (6, y) is 16.

Answer 1

\[d=\sqrt{(x-x1)^2+(y-y1)^2}.......16^2=(-5^2)+(10-y)^2\]
\[y^2-20y-131=0\]

Answer 2

draw a right angled triangle and use the pythagorean theorem. One of the sides is 6-1, the other is 10-y. The hypotenuse is 16.

so something like \[(6-1)^{2}+(10-y)^{2}=16^{2}\]

Answer 3

\[y=10+\sqrt{231} =25.1987 \]