Find all points that are 13 units away from the point (2,7) and that lie on the line x-2y=10.
Write your answer as a list of ordered pairs, separated by commas. If there are no such points, write "none".
Hint(s):
What shape is the graph of all the points that are 13 away from (2,7)? And what shape is the graph of the equation in the problem? At how many points might we expect those two shapes to intersect?
The equation of the line is x−2y=10. We can rewrite this equation as y=21x−5.
The distance formula is d=sqrt(x1−x2)2+(y1−y2)2.
Let (x,y) be a point on the line x−2y=10 that is 13 units away from the point (2,7).
Then d=sqrt(x−2)2+(y−7)2=13.
Substituting the equation y=21x−5 into the distance formula, we get:
(x−2)2+(21x−5−7)2=13
(x−2)2+(21x−12)2=13
x2−4x+4+x2−6x+36=169
2x2−10x+40=169
2x2−10x−129=0
We can factor this equation as:
(2x−17)(x+7)=0
Therefore, x=17/2 or x=−7.
If x=17/2, then y=21x−5=17/4−5=−3/4.
If x=−7, then y=21x−5=−27−5=−17/2.
Therefore, the two points that are 13 units away from the point (2,7) and that lie on the line x−2y=10 are (17/2,−3/4) and (−7,−17/2).
