SOLUTION: Find the point with coordinates of the form (a, 3a) that is in the third quadrant and is a distance 5 from P(2, 1)...

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Question 994253: Find the point with coordinates of the form (a, 3a) that is in the third quadrant and is a distance 5 from
P(2, 1)...
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Look for the negative value for a.

Distance Formula, sqrt%28%28a-2%29%5E2%2B%283a-1%29%5E2%29=5


a%5E2-4a%2B4%2B9a%5E2-6a%2B1=25

10a%5E2-10a%2B5=25

10%28a%5E2-a%29=20

a%5E2-a=2

a%5E2-a-2=0

%28a%2B1%29%28a-2%29=0
a=-1 will put the coordinates into quadrant #3, to be point (-1,-3).

The other solution of a=2 will not be in quadrant #3.