Locus need help.. (1 Viewer)

[kbei]

Ok, I have no idea how to do the following questions

First, fine the equation of the locus of a point P(x,y) that is subject to the following conditions:

1.)A point is √2 units from the line equation y=x. The answer is y=x plus or minus 2. However, I have no idea how that is right.

2.) A and B are fixed points (a,o) and (-a,o). Find the locus of P(x,y) such that the gradient of AP is twice the gradient of BP.

Any help along with some explanations would help, thanks heaps in advance.

 

[shaon0]

Ok, I have no idea how to do the following questions

First, fine the equation of the locus of a point P(x,y) that is subject to the following conditions:

1.)A point is √2 units from the line equation y=x. The answer is y=x plus or minus 2. However, I have no idea how that is right.

2.) A and B are fixed points (a,o) and (-a,o). Find the locus of P(x,y) such that the gradient of AP is twice the gradient of BP.

Any help along with some explanations would help, thanks heaps in advance.

1. For this question. Think about it like this. (If your a visual thinker, it may help drawing a diagram.) What points are their such that, the points are always 2 units away from y=x ie. the lines are two lines with the same gradient as y=x but different x and y-intercepts.

2. For this question it always helps using the distance formula and if necessary the point gradient form.

Sorry i couldn't help anymore.

 

[clintmyster]

with 2 i had a shot. Not 100% sure its right though

iight you know that 2 x gradient of BP = Gradient AP

therefore 2y / (x+a) = y / (x - a)

upon cancelling the y's

2x - 2a = x + a

x = 3a

 

[lyounamu]

Ok, I have no idea how to do the following questions

First, fine the equation of the locus of a point P(x,y) that is subject to the following conditions:

1.)A point is √2 units from the line equation y=x. The answer is y=x plus or minus 2. However, I have no idea how that is right.

2.) A and B are fixed points (a,o) and (-a,o). Find the locus of P(x,y) such that the gradient of AP is twice the gradient of BP.

Any help along with some explanations would help, thanks heaps in advance.

1) use the perpendicular distance formula:

sqrt(2) = abs(x - y)/(sqrt(2)
2 = abs(x-y)
x-y = 2 or y -x = 2
y = x-2 or y = x+2

2)
(y-0)/(x-a) = 2(y-0)/(x+a)
y/(x-a) = 2y/(x+a)
y(x+a) = 2y(x-a)
yx + ya = 2yx - 2ay
yx - 3ay = 0
y(x - 3a) = 0
x = 3a or y = 0

 

[Trebla]

Ok, I have no idea how to do the following questions

First, fine the equation of the locus of a point P(x,y) that is subject to the following conditions:

1.)A point is √2 units from the line equation y=x. The answer is y=x plus or minus 2. However, I have no idea how that is right.

2.) A and B are fixed points (a,o) and (-a,o). Find the locus of P(x,y) such that the gradient of AP is twice the gradient of BP.

Any help along with some explanations would help, thanks heaps in advance.

1) is a bit vague. Does it mean √2 units from the line by the shortest distance?
In that case: √2 = | x - y | / √2 by perpendicular distance formula
=> | x - y | = 2
So
x - y = 2 => y = x - 2
OR
x - y = -2 => y = x + 2

2) Gradient of AP: y / (x - a)
Gradient of BP: y / (x + a)
Thus:
y / (x - a) = 2y / (x + a)
(x + a)y = 2y(x - a)
xy + ay = 2xy - 2ay
xy = 3ay
y(x - 3a) = 0
This gives x = 3a IF y is not zero.
If y = 0, then you have the rather trivial case of the x-axis, which implies the gradient of AP and BP would be zero lol, but the condition would still hold.

 

[kbei]

er. yeah I checked these answers to the answers provided and they matched, I just wasnt sure how to work them out. I only just realised that the perpendicular point formula could be applied to the first question and the second question was pretty straighforward once I realised how to start it (by reading your posts ). Anyways thanks heaps.