help again geometic locus: r: y=-4/3x + 3 and s: y=-1

hellawowser

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Consider the family of all circumferences that are simultaneously tangent to the lines r: y=-4/3x + 3 and s: y=-1. What is the geometric locus and its equation formed by all the circumferences center?

I think it is a line but how do i get to the equation??
 
Have you considered constructing it?

What do circles tangent to a given line look like?
 
Have you considered constructing it?

What do circles tangent to a given line look like?
I did, like i said it matters only the center of the circumferences, which clearly looks like a line.

geogebra.jpg

I've done more simplier exercices before that given a circumference you should find the line tangent to certain point but that looks beyond.
 
Do you know that a radial line from the center of a circle to a tangent point is perpendicular to the tangent line?

Note that the right triangles share two legs with one another.

What does this say about the relationship between the line you seek and the acute angle formed by the two given lines?

locus.JPG
 
Do you know that a radial line from the center of a circle to a tangent point is perpendicular to the tangent line?

Note that the right triangles share two legs with one another.

What does this say about the relationship between the line you seek and the acute angle formed by the two given lines?

View attachment 8553

So the line is the bisector angle? I will try to find a point (center) of a circle that matches the two lines and then apply it to the line equation using the half of angle between them(problem is that my teacher hasn't taught that,so i don't know if i supposed to use this,but its very easy though). I will see what happens. Thanks
 
I did, like i said it matters only the center of the circumferences, which clearly looks like a line.

View attachment 8552

I've done more simplier exercices before that given a circumference you should find the line tangent to certain point but that looks beyond.

That's excellent. Next time, start with that. :)
 
That's excellent. Next time, start with that. :)

No need to deal with angles. Found a easier way. As of the distance between a the center and both lines are equal(radius),i just needed to equal both distance equations.Just be careful with the module on either side,there are both situations:they are both positive or both negative(eliminate module) or they positive and negative(eliminate module by multiplying one side (-1). So you get two equations,which are the lines of the centers.y=2x-7 and y=-1/2x+1/2

4a.jpg

Thanks for all the help
 
No need to deal with angles. Found [an] easier way …

y=2x-7 and y=-1/2x+1/2
Good work! (I'm not familiar with the module terminology, but your answers agree with mine.)

I didn't intend to imply that you had to calculate angle measurements (if that's what you you were thinking). I simply wanted you to realize you were seeking lines with slope that bisected the acute (and its supplementary) angle.

My way (for the acute-angle bisector):

slope = tan[1/2*arctan(-4/3)] = -1/2

Using Point-Slope Form -- we're free to do this, as the instructions do not constrain -- the answer is:

y + 1 = -1/2*(x - 3)

The other, perpendicular slope is 2, by property.

"Easier" is subjective. ;) ;)
 
Good work! (I'm not familiar with the module terminology, but your answers agree with mine.)

Yea sorry english is not my native language . I think Absolute value/Modulus are the right terms.
Transformed y=-4/3x + 3 to 4x+3y-9=0

|4x+3y-9|/ Square root 4^2 + 3^2 = |y+1|
 
Beer soaked praise follows.
No need to deal with angles. Found a easier way. As of the distance between a the center and both lines are equal(radius),i just needed to equal both distance equations.Just be careful with the module on either side,there are both situations:they are both positive or both negative(eliminate module) or they positive and negative(eliminate module by multiplying one side (-1). So you get two equations,which are the lines of the centers.y=2x-7 and y=-1/2x+1/2

View attachment 8555

Thanks for all the help
Very creative!
I wish all students are as imaginative as you are.
I see a great future waiting for you.

P.S. What software are you using to generate that wonderful graph?
 
Beer soaked comment follows.
My way (for the acute-angle bisector):

slope = tan[1/2*arctan(-4/3)] = -1/2

Using Point-Slope Form -- we're free to do this, as the instructions do not constrain -- the answer is:

y + 1 = -1/2*(x - 3)
I have the same thing in mind yesterday but I guess I forgot to look for an easier way out.
Clever kid that hellawowser.
 
Beer soaked afterthought follows.
I guess I forgot to look for an easier way out.
Scratch that.
A lot more work involved than I thought than just using slope = tan[1/2*arctan(-4/3)] = -1/2
A unit circle with center at (3,-1).
One point at (2,-1) and the other at intersection of y=-4/3x + 3 and upper half portion of said unit circle. Midpoint of those two points and then slope between midpoint to circle center.
That's a lot of work for just a slope.
 
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