SOLUTION: find the co-ordinates of the points on the line X+5Y=13 at a distance of 2 units from the line 12X-5Y+26=0.

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Question 675318: find the co-ordinates of the points on the line X+5Y=13 at a distance of 2 units from the line 12X-5Y+26=0.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I hope you have seen the formula for directed distance from a point to a line:
For the directed distance from a point, (x%5B1%5D, y%5B1%5D) to a line whose equation is in the form: Ax + By + C = 0, the distance is:
d+=+%28Ax%5B1%5D+%2B+By%5B1%5D+-+abs%28C%29%29%2Fsqrt%28A%5E2%2BB%5E2%29 if C < 0
or
d+=+%28-Ax%5B1%5D+-+By%5B1%5D+-+abs%28C%29%29%2Fsqrt%28A%5E2%2BB%5E2%29 if C > 0

Since the "C" of 12x -5y +26 = 0 is positive, we will be using the second form.

This formula gives a "directed" distance. This means that points on one side of the line are considered to have positive distances from the line and points on the other side of the line are considered to have negative distances. Since we do not care which side on the line the points we are looking for are located we will use 2 and -2 for the "d" so we can find points on either side of the line that at a distance of 2 from the line.

Substituting in the A, B and C from the equation and the 2 and -2 for "d" we get the following equations:

and


Simplifying the first equation...

2+=+%28-12x%5B1%5D+%2B+5y%5B1%5D+-+26%29%2Fsqrt%28144%2B25%29
2+=+%28-12x%5B1%5D+%2B+5y%5B1%5D+-+26%29%2Fsqrt%28169%29
2+=+%28-12x%5B1%5D+%2B+5y%5B1%5D+-+26%29%2F13
26+=+-12x%5B1%5D+%2B+5y%5B1%5D+-+26
Using similar steps on the second equation (with -2 for "d") we should get:
-26+=+-12x%5B1%5D+%2B+5y%5B1%5D+-+26

These two equations are equations of lines. Specifically they are equations of lines that are parallel to 12x -5y +26 = 0 and are at a distance of 2 from 12x -5y +26 = 0. Now we want to find the points where these lines intersect the line x + 5y = 13. To find these points I will use the Substitution Method (since I think that will be easier). Solving x + 5y = 13 for x we get:
x = -5y + 13
Substituting this expression for x into one of our parallel lines we get:
26+=+-12%28-5y+%2B+13%29+%2B+5y+-+26
Solving for y:
26+=+60y+%2B+%28-156%29+%2B+5y+-+26
26+=+65y+%2B+%28-182%29
208+=+65y
208%2F65+=+y
which reduces to:
16%2F5+=+y
With this value for y we can find the x coordinate using the x = -5y + 13 equation:
x+=+-5%2816%2F5%29+%2B+13
x+=+-16+%2B+13
x+=+-3
So one of the points we are looking for is (-3, 16/5).

Substituting for x in the equation for the second parallel line (where d = -2) we get:
-26+=+-12%28-5y+%2B+13%29+%2B+5y+-+26
Solving for y:
-26+=+60y+%2B+%28-156%29+%2B+5y+-+26
-26+=+65y+%2B+%28-182%29
156+=+65y
156%2F65+=+y
which reduces to
12%2F5+=+y
Now we can find this y value's x-coordinate:
x+=+-5%2812%2F5%29+%2B+13
x+=+-12+%2B+13
x+=+1
So the other point we are looking for is (1, 12/5)

So there are two points on the line x + 5y = 13 that are at a distance of 2 from the line 12x + 5y + 26 = 0: (-3, 16/5) and (1, 12/5)