Locus equation (1 Viewer)

[davidbarnes]

Can anyone please give me a rundown of how to do to the following question:

"Find the equation in exact form of the locus of a point that moves so that its distance from the line x - 2y - 3 = 0 is always 5 units."

The line x - 2y - 3 = 0 being in there has just confused me.

 

[ssglain]

Can anyone please give me a rundown of how to do to the following question:

"Find the equation in exact form of the locus of a point that moves so that its distance from the line x - 2y - 3 = 0 is always 5 units."

The line x - 2y - 3 = 0 being in there has just confused me."

Whenever a question talks about the distance between a point and a line, this refers to the perpendicular distance. All you need to do is let the point be P(x, y) and put this in the perp dist formula. Have a go for yourself. the solution is under the spoiler.

Spoiler

Formula: perpendicular distance = |ax + by + c|/(a² + b²)^1/2 where a, b, c are coefficients of the equation of line and x, y are the co-ordinates of the point. i.e. a = 1, b = -2, c = -3

Hence 5 = |x - 2y - 3|/{1² + (-2)²}^1/2
5*(5)^1/2 = x - 2y - 3
.: Locus: x - 2y +2*(5)^1/2 = 0

 

[getridofns]

You should first draw a diagram to see what this means. Just rought diagram. So for P to remain a distance of 5 units from the line, it must move in a straight line. After you identify this you can the proceed with what ssglain done.




Try this one.

A point P(x,y) moves so that its distance from the origin is always 5units. Try to draw a diagram

 

[SoulSearcher]

Can anyone please give me a rundown of how to do to the following question:

"Find the equation in exact form of the locus of a point that moves so that its distance from the line x - 2y - 3 = 0 is always 5 units."

The line x - 2y - 3 = 0 being in there has just confused me."

Whenever a question talks about the distance between a point and a line, this refers to the perpendicular distance. All you need to do is let the point be P(x, y) and put this in the perp dist formula. Have a go for yourself. the solution is under the spoiler.

Spoiler

Formula: perpendicular distance = |ax + by + c|/(a² + b²)^1/2 where a, b, c are coefficients of the equation of line and x, y are the co-ordinates of the point. i.e. a = 1, b = -2, c = -3

Hence 5 = |x - 2y - 3|/{1² + (-2)²}^1/2
5*(5)^1/2 = x - 2y - 3
.: Locus: x - 2y +2*(5)^1/2 = 0

You also have to recognise that in this particular type of question there are actually two solutions that can fit the parameters of what the question is asking for. So here, you would also have the solution where 5.5^0.5 = -(x - 2y - 3)

 

[ssglain]

You should first draw a diagram to see what this means. Just rought diagram. So for P to remain a distance of 5 units from the line, it must move in a straight line. After you identify this you can the proceed with what ssglain done.

You also have to recognise that in this particular type of question there are actually two solutions that can fit the parameters of what the question is asking for. So here, you would also have the solution where 5.5<SUP>0.5</SUP> = -(x - 2y - 3)

These are great points. I overlooked the fact that another locus also satisfies the parameters. It's a good thing SS pointed out. I'll make all my mistakes now and hope to avoid them in the HSC.

 

[SoulSearcher]

These are great points. I overlooked the fact that another locus also satisfies the parameters. It's a good thing SS pointed out. I'll make all my mistakes now and hope to avoid them in the HSC.

I know I'll make mistakes in the HSC exams, I've just got to minimise the damage now

 

[SoulSearcher]

r^2 = (x-a)^2 + (y-b)^2 + (z-c)^2

No spheres please.

 

[Ahmed A]

Isnt it simply...
PA = 5PB
where P(x,y) x-2y-3=0
x=0, y=-3/2
y=0, x=3
therefore, A(0,-3/2), B(3,0)
PA=5PB
therefore the distance of PA= 5 x (the distance of PB)
=
(square root of)[(x-0)^2- (y- -3/2)^2] = 5(square root of)[(x-3)^2-(y-0)^2]
=
so on...
...???

 

[ssglain]

lets do locus in 3D!

3D co-ordinate geometry is one thing they should be introducing into the new maths syllabus - for MX2 at least. Think how much more fun that will be.

 

[ssglain]

Isnt it simply...
PA = 5PB
where P(x,y) x-2y-3=0
x=0, y=-3/2
y=0, x=3
therefore, A(0,-3/2), B(3,0)
PA=5PB
therefore the distance of PA= 5 x (the distance of PB)
=
(square root of)[(x-0)^2- (y- -3/2)^2] = 5(square root of)[(x-3)^2-(y-0)^2]
=
so on...
...???

The "distance" here refers to the perpendicular distance. The variable point P(x, y) is required to be perpendicularly 5 units away from the given line. The parameters do not relate to any specific points on the line such as the intercepts. I'm not sure why you needed to find the points A and B. You seem to have misread the question. The question did not ask for the locus of a point P (x, y) such that its distance from the y-intercept of the line x - 2y - 3 = 0 is five times its distance from the x-intercept of the same line. Maybe try to read the question again. My solution is there, but I missed the equally valid locus lying below x - 2y - 3 = 0 as SS pointed out.

 

[Ahmed A]

The "distance" here refers to the perpendicular distance. The variable point P(x, y) is required to be perpendicularly 5 units away from the given line. The parameters do not relate to any specific points on the line such as the intercepts. I'm not sure why you needed to find the points A and B. You seem to have misread the question. The question did not ask for the locus of a point P (x, y) such that its distance from the y-intercept of the line x - 2y - 3 = 0 is five times its distance from the x-intercept of the same line. Maybe try to read the question again. My solution is there, but I missed the equally valid locus lying below x - 2y - 3 = 0 as SS pointed out.

OH YEAHH.....
totally misread the question my bad....