# Figuring out X on a parallel line without graphing

• Oct 25th 2007, 07:19 PM
simplekin
Figuring out X on a parallel line without graphing
Hello, i'm stuck on this question:

Determine the value of x so that the line segment with endpoints A(x,3) and B(1,7) is parallel to the line segment with endpoints C(2,-4) and D(5,-2)

Ok, i got the slope by using the forumla Y2 -Y1 over X2 - X1

-4 -(-2)
---------
2 -5
=
-2
-----
-3

That is the slope, now everytime i get to this point, i have no choice but to multiply, so in this case X = -5, so coordinates would be (-5, 3). But this answer was found through graphing. On a test, i won't have the time to graph, so how can i find x without having to graph it (when i have the slope already).
• Oct 25th 2007, 07:49 PM
Jhevon
Quote:

Originally Posted by simplekin
Hello, i'm stuck on this question:

Determine the value of x so that the line segment with endpoints A(x,3) and B(1,7) is parallel to the line segment with endpoints C(2,-4) and D(5,-2)

Ok, i got the slope by using the forumla Y2 -Y1 over X2 - X1

-4 -(-2)
---------
2 -5
=
-2
-----
-3

That is the slope, now everytime i get to this point, i have no choice but to multiply, so in this case X = -5, so coordinates would be (-5, 3). But this answer was found through graphing. On a test, i won't have the time to graph, so how can i find x without having to graph it (when i have the slope already).

to be parallel means they have the same slope, so using the $\displaystyle \frac {y_2 - y_1}{x_2 - x_1}$ formula, we want:

$\displaystyle \frac {7 - 3}{1 - x} = \frac {-2 - (-4)}{5 - 2}$

now solve for $\displaystyle x$