Find all points (x, y) that are 13 units away from the point (2,7) and that lie on the line x-2y=10 ​

(14,2) and (-6/5,-28/5)

Step-by-step 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).