Question 812925
Since the problem talks about distance between points in the coordinate plane we will be using the distance formula:
{{{d = sqrt((x[2]-x[1])^2 + (y[2]-y[1])^2)}}}<br>
We are given everything but the y coordinate(s) of the desired points. With only one unknown in the formula we can use it find that unknown:
{{{13 = sqrt((3-(-2))^2 + (y[2]-(-1))^2)}}}
We start by simplifying:
{{{13 = sqrt((3+2)^2 + (y[2]+1)^2)}}}
{{{13 = sqrt(25 + y[2]^2+2y[2]+1)}}}
{{{13 = sqrt(y[2]^2+2y[2]+26)}}}
Now we square both sides:
{{{169 = y[2]^2+2y[2]+26}}}
This is a quadratic so we want a zero on one side:
{{{0 = y[2]^2+2y[2]-143}}}
Now we factor:
{{{(y[2]+13)(y[2]-11)=0}}}
From the Zero Product Property:
{{{y[2]+13 = 0}}} or {{{y[2]-11=0}}}
Solving:
{{{y[2] = -13}}} or {{{y[2]= 11}}}
So there are two points with an x-coordinate of 3 which are 13 units away from (-2, -1): (3, -13) and (3, 11)