Answer:
(14,2) and (-6/5,-28/5)
Explanation:
The distance, d, from two points (x,y) and another point (a,b) can be calculated using
d=sqrt((x-a)^2+(y-b)^2).
Our point (a,b) is (2,7) and d=13.
Making substitutions:
13=sqrt((x-2)^2+(y-7)^2)
We are also given the relation between x and y is given as x-2y=10.
Adding 2y to both sides gives: x=10+2y
Make this insertion into our equation:
13=sqrt((10+2y-2)^2+(y-7)^2)
Simplify inside:
13=sqrt((8+2y)^2+(y-7)^2)
Square both sides:
169=(8+2y)^2+(y-7)^2
Expand binomial squares:
169=64+32y+4y^2+y^2-14y+49
Combine like terms:
169=5y^2+18y+113
Subtract 169 on both sides:
0=5y^2+18y-56
We could try to factor
0=(5y+28)(y-2)
So y=2 or y=-28/5
Recall x=10+2y
So if y=2, then x=10+2(2)=10+4=14.
So if y=-28/5, then x=10+2(-28/5)=10+(-56/5)
=50/5 +-56/5
=-6/5.
So two points satisfying given criteria is
(14,2) and (-6/5,-28/5).