SOLUTION: Find all points on the y-axis (that is, having coordinates (0,y), where y is to be determined) that are 6 units distant from the point (4, -3). Thank you in advance

Algebra ->  Finance -> SOLUTION: Find all points on the y-axis (that is, having coordinates (0,y), where y is to be determined) that are 6 units distant from the point (4, -3). Thank you in advance      Log On


   



Question 775724: Find all points on the y-axis (that is, having coordinates (0,y), where y is to be determined) that are 6 units distant from the point (4, -3). Thank you in advance
Found 2 solutions by josgarithmetic, psbhowmick:
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
Shortest distance to the y axis is what is helpful. No change in y value, from y=-3. You would form a triangle. One point is the given point (4, -3). Another point is (0, -3) which is the distance TO the y axis. The last point is ON the y axis but 6 units from (4, -3). You just need the Distance Formula.

In fact, you might try this with just Pythagorean Theorem, and then avoid formal use of Distance formula. You have two lengths. One is 6 units, the other is 3 units. Size of the other leg?
L%5E2%2B3%5E2=6%5E2, since the hypotenuse is 6 units.
L%5E2=36-9
L%5E2=25
L=5 units long, but really, L=%2B5 or L=-5.

Up 5 units from -3 is -3+5=2.
Down 5 units from -3 is -3-5=-8.

The answer for the point is then either or both of (0,2) or (0,-8).

Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
Let a point on y-axis whose distance from (4,-3) is 6 units be having coordinates (0,k).

The distance between these two points must be 6 units.

%284+-+0%29%5E2+%2B+%28-3+-+k%29%5E2+=+6%5E2
4%5E2+%2B+%28k%2B3%29%5E2+=+6%5E2
16+%2B+%28k%2B3%29%5E2+=+36
%28k%2B3%29%5E2=20
k%2B3=+0%2B-+2%2Asqrt%285%29
k+=+-3+%2B-+2%2Asqrt%285%29

Therefore there are two such points: (0,-3+%2B+2%2Asqrt%285%29) and (0,-3+-+2%2Asqrt%285%29)