Thread: Help with a problem involving distance and midpoint?

Hello everyone. I tried solving this problem (attached photo) and arrived at the answer k=±4 for number 1. I tried solving number 2 but I just don't get how to do it. Any help would be very much appreciated. Thanks a bunch!

You want $|y|=|x|+1$ and you are given the point $(1,k)$. So, plugging in, you have

$$|k|=|1|+1$$

This gives $|k|=2$ and $k=\pm 2$.

Point D is (0, 2) and point Y is (2, -2). The midpoint of interval DY is ((0+ 2)/2, (2- 2)/2)= (1, 0). A is the point (1, k). The square of the distance between A and (1, 0) is $\displaystyle (1- 1)^2+ (k- 0)^2= k^2= 4^2= 16$. What are the solutions to $\displaystyle k^2= 16$?

Originally Posted by HallsofIvy

Point D is (0, 2) and point Y is (2, -2). The midpoint of interval DY is ((0+ 2)/2, (2- 2)/2)= (1, 0). A is the point (1, k). The square of the distance between A and (1, 0) is $\displaystyle (1- 1)^2+ (k- 0)^2= k^2= 4^2= 16$. What are the solutions to $\displaystyle k^2= 16$?

The OP said that he/she solved that problem. He/she even stated the answer of $\pm 4$.