Question 147055: find the shortest distance between the parallel lines with equations 5x-12y+33=0 and 5x-12y-6=0
PLEASE HELP!!!
Found 2 solutions by edjones, bucky:
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! 5x-12y+33=0
5x-12y-6=0
.
5x-12y+33=0
-12y=-5x-33
y=5x/12 + 33/12
.
5x-12y-6=0
-12y=-5x+6
y=5x/12 - 1/2
.
The distance between the parallel lines is |33/12 + 1/2|=3.25
This is easier to see if we change the slope from 5x/12 (1st graph) to 0x/12 (2nd graph)
.
Ed
.
.
Answer by bucky(2189)  (Show Source):
You can put this solution on YOUR website! The answer to this problem is 3, not 3.25
.
The shortest distance between the parallel lines is the length of the perpendicular
between the two lines, not the distance between the two intercepts on the y-axis.
.
Do it this way:
.
(1) Convert the two equations to slope intercept form. When you do, you will get:
.
y = (5/12)x + (33/12) and
.
y = (5/12)x - (6/12)
.
Since the perpendicular to a line has a slope that is the negative inverse of the slope
of the given line, the first slope intercept equation has a perpendicular of the form:
.
y = (-12/5)x + (+33/12)
.
and it also crosses the y-axis at the point (0, 33/12)
.
The question then becomes at what point does this perpendicular line cross the second
given line. Find this by simultaneously solving the second equation and the perpendicular
equation. In other words solve the equation pair:
.
y = (5/12)x - (6/12) and
y = (-12/5)x + (33/12)
.
You should find that the common solution of this pair is the coordinate point (15/13, -1/52)
.
Now you have the two critical points where the perpendicular crosses both of the graphs
of the original equations. The graph of what you have is shown below:
.
.
The point where the purple perpendicular line crosses the red graph is (0,33/12) and
where it crosses the green line is (15/13, -1/52).
.
Now just use the distance formula to calculate the distance between these two points, and
you will find that the answer is 3.
.
Hope this clarifies things for you.
.
|
|
|
