maths help (1 Viewer)

[bubblesss]

Re: maths again

Perpindicular distance formula with x - y = 0 with P(x,y) and equate it to sqrt2.

if you want the answer, just reply again.

could u please show me the working?

 

[tommykins]

Re: maths again

a point is square root 2 units from the line with the equation y=x. find the locus of the point? forgot how to do it.thanks in advance.

P (x,y)
x - y = 0

a = 1
b = -1
c =0

So we have D = |(ax + by + c)/sqrt(a²+b²)|

sqrt2 = |x - y/sqrt(1²+[-1]²)| = |x-y/sqrt2|

Square both sides to get rid of the abosolute value.

2 = (x-y)²/2
4 = (x-y)² is your locus.

 

[bubblesss]

Re: maths again

the distance of a point from the line y=-3 is two fifths of its distance from the line y=-1. find the locus of the point?
PA^2 =2/5 PB^2
do we square the 2/5 as well?

 

[tommykins]

Re: maths again

Yes, you sqaure the 2/5 as well. This is a crucial step, DO NOT FORGET IT.

You probably don't need to do the algebra in that though, but a hint would be to graph it and look at the question

If you're stuck, ill post up solutions.

 

[tacogym27101990]

Re: maths again

so you have

|x-y| = rt2
rt2

|x-y|=2

so x-y=2 or x-y=-2

so the locus is the lines x-y-2=0 and x-y+2=0

 

[bubblesss]

Re: maths again

Yes, you sqaure the 2/5 as well. This is a crucial step, DO NOT FORGET IT.*You probably don't need to do the algebra in that though, but a hint would be to graph it and look at the question *If you're stuck, ill post up solutions.

*thanks
but if i square it im not getting the answer? im stuck!!!

 

[bubblesss]

Re: maths again

so you have

|x-y| = rt2
rt2

|x-y|=2

so x-y=2 or x-y=-2

so the locus is the lines x-y-2=0 and x-y+2=0

uhh what formula is that? sry for bugging u

 

[Dr. Zoidberg]

Re: maths again

http://img329.imageshack.us/img329/9115/mathsshz0.jpg

 

[tacogym27101990]

Re: maths again

uhh what formula is that? sry for bugging u

thats just the perp. distance formula

but its for the 1st question

youd just asked another question by the time i wrote it out

 

[tommykins]

Re: maths again

the distance of a point from the line y=-3 is two fifths of its distance from the line y=-1. find the locus of the point?
PA^2 =2/5 PB^2
do we square the 2/5 as well?

P(x,y)

sqrt[ (x-x)² + (y+3)²] = 2/5sqrt[ (x-x)² + (y+1)² ]

(y+3)² = 4/25(y+1)²
25(y+3)² = 4(y+1)²
25y² + 150y + 225 = 4y²+8y + 4

21y² + 142y + 221 = 0 is your locus.

Could solve for y, but neh.

 

[bubblesss]

Re: maths again

P(x,y)

sqrt[ (x-x)² + (y+3)²] = 2/5sqrt[ (x-x)² + (y+1)² ]

(y+3)² = 4/25(y+1)²
25(y+3)² = 4(y+1)²
25y² + 150y + 225 = 4y²+8y + 4

21y² + 142y + 221 = 0 is your locus.

Could solve for y, but neh.

uhh the answer is 5x - 2y +13=0 and 5x +2y+17=0

dont wry i'll try to work sumthing out.

 

[tommykins]

Re: maths again

uhh the answer is 5x - 2y +13=0 and 5x +2y+17=7

dont wry i'll try to work sumthing out.

Can't be, the lines you gave me were y = -3, y = -1.

If a point is always 2/5 distance away from those 2 lines, then it must be another y line.

 

[SeftonIsAHole]

Re: maths again

P (x,y)
x - y = 0

a = 1
b = -1
c =0

So we have D = |(ax + by + c)/sqrt(a²+b²)|

sqrt2 = |x - y/sqrt(1²+[-1]²)| = |x-y/sqrt2|

Square both sides to get rid of the abosolute value.

2 = (x-y)²/2
4 = (x-y)² is your locus.

instead of squaring both sides cant you do take the sqrt2 up so | x- y| = 2
so the locus is x-y = 2 or x-y=-2 not (x-y)^2 = 4
.. correct me if i'm wrong.

 

[bubblesss]

Re: maths again

Can't be, the lines you gave me were y = -3, y = -1.

If a point is always 2/5 distance away from those 2 lines, then it must be another y line.

oops sry it was x=-3
my bad
but still i got 25x^2 - 4y^2 + 150x-8y+221=0
nowhere near the real answer!!!!!!!1

 

[tommykins]

Re: maths again

instead of squaring both sides cant you do take the sqrt2 up so | x- y| = 2
so the locus is x-y = 2 or x-y=-2 not (x-y)^2 = 4
.. correct me if i'm wrong.

Both yield the same answer, I just feel alot safer with my method. They're both the same graph.

the distance of a point from the line x=-3 is two fifths of its distance from the line y=-1. find the locus of the point?
PA^2 =2/5 PB^2

Let the P(x,y)
sqrt[ (x+3)² + (y-y)² ] = 2/5[ (x-x)² + (y+1)² ]
(x+3)² = 4/25(y+1)²

25(x+3)² = 4(y+1)²
25x² + 150x + 225 = 4y² + 8y + 4

25x² + 150x + 225 - 4y² - 8y - 4 = 0 is your locus.

 

[SeftonIsAHole]

Re: maths again

k thanks tommyskins
lol, dont you have better stuff to do like study for ur hsc, rather than answering math questions for people on bos :apig:

 

[tacogym27101990]

Re: maths again

instead of squaring both sides cant you do take the sqrt2 up so | x- y| = 2
so the locus is x-y = 2 or x-y=-2 not (x-y)^2 = 4
.. correct me if i'm wrong.

yeah either way works
your ways the way i did it

 

[tommykins]

Re: maths again

k thanks tommyskins
lol, dont you have better stuff to do like study for ur hsc, rather than answering math questions for people on bos :apig:

I can multi-task.

 

[bubblesss]

Immediate Help!!!!!!!!!11

please can sum1 solve my confusion by solving this?

the distance of a point from the line x = -3 is two fifths of its distance from the line y= -1. find the locus of the line. please show appropriate working. thanks.

 

[bubblesss]

can someone please help me on this locus question?
the distance of a point from the line x= - 3 is two fifths of its distance from the line y= -1. find the locus?