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=(x2-x1)²+(y2-y1)²}}}
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
<pre>You're on the right track, but you forgot to square the 15!
{{{15 = (x[2] - 10)^2 + 81}}} <-------- This is what you have
{{{15^2 = (x[2] - 10)^2 + 81}}} <------- This is what you should have
{{{225 = x[2]^2 - 20x + 100 + 81}}}
Continue to solve for {{{x[2]}}} from here. You should get 2 values for {{{x[2]}}}, but one is positive and the
other is negative. You need the negative value since K is in the 2<sup>nd</sup> quadrant, where x is negative.