Help with a problem involving distance and midpoint?

Sep 2018
27
0
Philippines
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! (Happy)
 

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SlipEternal

MHF Helper
Nov 2010
3,728
1,571
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 $k=\pm 2$.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
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 \(\displaystyle k^2= 16\)?
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
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 $\pm 4$.