Why is my Locus Wrong? (1 Viewer)

[Sy123]

The exact question:

The point P(x, y) is equidistant from the lines:



And



Find the locus of point P.


What I did:

Let variable point
A vary on x=3

Therefore the coordinates of A are:
A(3, y)

Let a variable point
B vary on the line 4x+3y-15=0

Therefore the coordinates of B are:



Lets take the distance formula of PA and PB, simplify all on to one side using our knowledge that PA=PB

Simplify, hence locus of P (I got a big expression with x^2 and y^2 and x and y etc.)

Apparently the locus is a striaght line(s)
I can see how this is, but the method you are supposed to do assumes perpendicular distance. If it said perpendicular distance then its easy and I know what to do, but the wording is vague and I want to know why my solution is mathematically incorrect (as far as I can tell, it definitely answers the question)

Thanks

 

[jenslekman]

i think you can't let A be (3, y) because the y value of P is already y and A and P do not share the same y values. if you get what i mean. try another method

 

[Sy123]

Well they are both part of the same planes arent they? :s
So I assume they would have the same y's?.....

for x=3, we dont know what y could be, it could be any value and when that varies along with P(x, y), we get equidistant?..

 

[jenslekman]

Well they are both part of the same planes arent they? :s
So I assume they would have the same y's?.....

for x=3, we dont know what y could be, it could be any value and when that varies along with P(x, y), we get equidistant?..

o i had anoth sol/n in mind

 

[Sy123]

Ok so when I graph this into geogebra:

That is:



You get two straight lines on the same plane, and they satisfy the equidistance between the two lines, if you were to use like perpendicular distance formula or something.
Does this mean Im right?

 

[Drongoski]

Locus = the 2 bisectors of the angles (1 acute, the other obtuse) between the 2 given lines.

 

[Peeik]

I cant but help to read the title as "Why is my name Lucas Wong?"

 

[Genero]

Generally, if the locus involves the distance between a point to a line, you would probably have to use perpendicular distance. There's really no other way by 2-unit methods that can conveniently compare the distance of a point to a line.

What you did was assumed a variable point that moves along the lines, then used the point P(x,y) with the distance formula and going PA = PB. The problem with that, I think, is that you'd have too many variables. The x of the line, the x of the point, and the x of the other line. X at one point on the line doesn't necessarily equal to the same x for the point, so I'd imagine you'd reach a dead end as there should only be 2 variables - the relationship between x and y. By making PA = PB, you've found a relationship between A(Xa, Ya), B(Xb, Yb) and P(Xp, Yp). I'm guessing you said all the X's and Y's were the same, hence coming up with an equation of sorts.

Not too sure though :L I'm just guessing that's how it is, though to be fair, you should probably use an easier method even if it is correct.

 

[Sy123]

Generally, if the locus involves the distance between a point to a line, you would probably have to use perpendicular distance. There's really no other way by 2-unit methods that can conveniently compare the distance of a point to a line.

What you did was assumed a variable point that moves along the lines, then used the point P(x,y) with the distance formula and going PA = PB. The problem with that, I think, is that you'd have too many variables. The x of the line, the x of the point, and the x of the other line. X at one point on the line doesn't necessarily equal to the same x for the point, so I'd imagine you'd reach a dead end as there should only be 2 variables - the relationship between x and y. By making PA = PB, you've found a relationship between A(Xa, Ya), B(Xb, Yb) and P(Xp, Yp). I'm guessing you said all the X's and Y's were the same, hence coming up with an equation of sorts.

Not too sure though :L I'm just guessing that's how it is, though to be fair, you should probably use an easier method even if it is correct.

Yeah well it seems I got it right, my relation between x and y is actually 2 straight lines in one.
Thanks anyway though

Im sure this thread has no use now