SOLUTION: Segment JK is exactly 15 units long. Point J is located at (10,-8). Point K is located in quadrant 2 and is one unit above the x-axis. What are the coordinates of point K? (May som

Algebra ->  Length-and-distance -> SOLUTION: Segment JK is exactly 15 units long. Point J is located at (10,-8). Point K is located in quadrant 2 and is one unit above the x-axis. What are the coordinates of point K? (May som      Log On


   

Question 1011570: Segment JK is exactly 15 units long. Point J is located at (10,-8). Point K is located in quadrant 2 and is one unit above the x-axis. What are the coordinates of point K? (May someone help me please and thank you)
* this is what I did before I confused myself:
After you add them together you have to find the square root, the square is the distance.(d stands for distance) I couldn't get the square root symbol I'm sorry.
d=%28x2-x1%29%B2%2B%28y2-y1%29%B2
d=15. 15=(x2-10)²+(y2-(-8))²
15=(x2-10)²+(9)²
15=(x2-10)²+(9)²
15=(x2-10)²+81
Found 2 solutions by josmiceli, MathTherapy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
You have the given point J( 10, -8 ). The signs ( +, - )
tell you it is in the 4th quadrant.
The other point K looks like ( x, 1 )
------------------------------
If you used J as a center and made part of a circle with
radius = 15 in the 2nd quadrant, it crosses the line
+y+=+1+ and that is point K
----------------------------
+15%5E2+=+%28+10+-+x%5B1%5D+%29%5E2+%2B+%28+-8+-+1+%29%5E2+
+225+=+100+-+20x%5B1%5D+%2B+x%5B1%5D%5E2++%2B+81+
+x%5B1%5D%5E2+-+20x%5B1%5D+%2B+181+-+225+=+0+
+x%5B1%5D%5E2+-+20x%5B1%5D+-+44+=+0+
I notice that +44+=+2%2A22+ so I can guess:
+%28+x%5B1%5D+-+22+%29%2A%28+x%5B1%5D+%2B+2+%29+=+0+
+x%5B1%5D+=+22+ doesn't work -it's in the 1st quadrant
+x%5B1%5D+=+-2+ gives me signs ( -, + ), which is 2nd quadrant
so that works
Point K is ( -2, 1 )
------------------
check answer:
+15%5E2+=+%28+10+-+x%5B1%5D+%29%5E2+%2B+%28+-8+-+1+%29%5E2+
+15%5E2+=+%28+10+-+%28-2%29+%29%5E2+%2B+%28+-8+-+1+%29%5E2+
+15%5E2+=+12%5E2+%2B+%28-9%29%5E2+
+225+=+144+%2B+81+
+225+=+225+
OK

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Segment JK is exactly 15 units long. Point J is located at (10,-8). Point K is located in quadrant 2 and is one unit above the x-axis. What are the coordinates of point K? (May someone help me please and thank you)
* this is what I did before I confused myself:
After you add them together you have to find the square root, the square is the distance.(d stands for distance) I couldn't get the square root symbol I'm sorry.
d=%28x2-x1%29%B2%2B%28y2-y1%29%B2
d=15. 15=(x2-10)²+(y2-(-8))²
15=(x2-10)²+(9)²
15=(x2-10)²+(9)²
15=(x2-10)²+81
You're on the right track, but you forgot to square the 15!

15+=+%28x%5B2%5D+-+10%29%5E2+%2B+81 <-------- This is what you have

15%5E2+=+%28x%5B2%5D+-+10%29%5E2+%2B+81 <------- This is what you should have

225+=+x%5B2%5D%5E2+-+20x+%2B+100+%2B+81

Continue to solve for x%5B2%5D from here. You should get 2 values for x%5B2%5D, but one is positive and the

other is negative. You need the negative value since K is in the 2nd quadrant, where x is negative.