# Locus help

1. Mar 22, 2005

### aisha

Locus help plz

Just started this unit and dont have a clue as to what is happening.

Here is the question

Describe the locus points that are 3 units from line y = 2x – 3. Give the equation of the locus. Draw a diagram as well.

Well I did a rough sketch of the line given and moved left and right 3 units of that line and I think the equations of those lines are y=2x+9 and y=2x-3

but how do I describe the locus? I dont understand what to do now?

2. Mar 23, 2005

### Andrew Mason

The locus is the set of points that satisfy those conditions. I think they want a more mathematical description: ie. set of all points (x,y) such that the length of the line (slope -1/2) passing through (x,y) from (x,y) and perpendicular to the line y = 2x - 3 is less than 3 units.

AM

Edit: In the last sentence 'less than' should be 'equal to'.

Last edited: Mar 23, 2005
3. Mar 23, 2005

### HallsofIvy

Staff Emeritus
The locus is, of course, the two lines on either side of the given line. They will be parallel to, and so have the same slope as, the given line, just as you have.

You say "I think the equations of those lines are y=2x+9 and y=2x-3". Hold on, wasn't y= 2x- 3 the given line??

Remember that the distance, 3, between lines is measured perpendicular to both lines, NOT vertically!

4. Mar 23, 2005

### aisha

oops I made a mistake the equations I have are

y=2x+1 , y=2x-3, and y=2x-7 What do I do now if these are correct?

Last edited: Mar 23, 2005
5. Mar 23, 2005

### Andrew Mason

How have you determined the distance from the line y=2x-3 to the other two? I think you have to work it out using the Pythagorean theorem.

Once you have the correct lines, the locus consists of the set of all points on the two lines that are 3 units distant from y=2x-3.

AM

6. Mar 23, 2005

### aisha

I got it y=2x and y=2x-6
2 parallel lines

7. Mar 24, 2005

### Andrew Mason

This does not result in the lines being 3 units apart in perpendicular distance.

Draw another line parallel to L (y = 2x-3); call it L1: y=2x + C.

Compare the y intercepts (x=0) of these two lines.

Choose C such that the vertical distance, d, between the intercepts is the hypotenuse of a right triangle. In this triangle, $dsin\theta = 3$ where $\theta$ is the angle between L and the y axis. You can see that the tangent of theta is the x intercept of L divided by the y intercept.

$$tan\theta = \frac{3/2}{-3} = -1/2$$

$$d = 3/sin\theta = 6.7082$$

This means the y intercept of L1 is 6.7082 units above the y intercept of L (which is -3). So the y intercept of L1 is 3.7082.

That means that C = 3.7082 so L1 is given by the equation y = 2x +3.7082

Similarly for L2 on the other side of L, the y intercept is 9.7082 so its equation is: y = 2x - 9.7082

Edit: .9082 should be .7082. I just thought it was a little easier using angles than using Pythagoras. Using Pythagoras, the distance along the y axis between the y intercepts of L and L1 is $3\sqrt{5}$ (ie. for right triangles from L to L1 with a vertical hypotenuse, the ratios of sides are 1, 2, $\sqrt{5}$. So the distance between y intercepts of L and L1, L2 is $3\sqrt{5}$

AM

Last edited: Mar 24, 2005
8. Mar 24, 2005

### aisha

I think you are getting two complicated my teacher said yes the slopes are the same the lines are parallel, why are you talking about perpendicular points? Even the equtions you gave are not perpendicular perpenicular would have been if there was -1/2 as the slope of the two new locus, Im confused on what your doing. I dont know why you got into theta when Im done the trig unit sorry for the confusin but thanks for all the help

9. Mar 24, 2005

### Data

The slopes are the same and the lines are parallel, correct (these two statements are equivalent, of course!).

The problem is that the points on your lines, $y = 2x$ and $y = 2x - 6$, are not a distance of 3 units from the line $y = 2x-3$. The y-intercepts are wrong. Graph the lines and see if you can tell why.

10. Mar 24, 2005

### Andrew Mason

The equations I gave are the parallel lines that are 3 units from the line y = 2x-3.

Ok forget about the angle approach and just use Pythagoras. Follow my explanation and draw the lines on the graph.

You will see that for lines with slope 2 that are 3 units apart in perpendicular distance are $3\sqrt{5}$ units apart vertically. This means that their y intercepts will differ by $3\sqrt{5}$. So if the y intercept of the line is -3, the lines that are 3 units in perpendicular distance from it will intersect the y axis a further $3\sqrt{5}$ units above or below it. The lines with slope 2 having those y intercepts will be the locus of the solution.

AM

11. Mar 24, 2005

### aisha

my teacher said that my answers are correct, I am still not sure why you think they are wrong I guess it doesnt matter now as long as I get the marks. Thanks for your help

12. Mar 24, 2005

### Andrew Mason

What is the width of the strip between the outside lines? Shouldn't it be 3+3 units? Is it?

I was just trying to help you understand the problem. If your teacher says that your answers are right, you should show her that she is wrong. If you do, you will get even more marks.

AM

13. Mar 24, 2005

### aisha

I dont know to solve this problem I just moved each point 3 down and 3 up keeping the slope the same and getting a new y intercept.

14. Mar 24, 2005

### Data

Here is a picture demonstrating the flaw in your solution. Show it to your teacher. The green line is $y = 2x - 3$, the brown line is $y = 2x$, and the yellow line is $y=2x - 6$.

The blue line is perpendicular to all three of them. For the correct answer to the question, the distance along the blue line from the green line to the brown/yellow lines should be 3. It is not.

The vertical distance at a particular $x$ from the green line to the brown/yellow lines is 3, but this is not what the question asks for.

The diagram is completely to scale, by the way, so you can see the problem very clearly. Here is the link:

http://www.geocities.com/data.8@rogers.com/Plot1.jpg