Locus. (1 Viewer)

[shaon0]

Given the points A(1,4) and B(-3,2) find the equation of the locus such that angle APB is a right angle.

 

[lolokay]

didn't we solve pretty much that exact question in the other thread?

 

[shaon0]

didn't we solve pretty much that exact question in the other thread?

that was with straight lines. this is locus.

 

[tommykins]

回复: Locus.

Given the points A(1,4) and B(-3,2) find the equation of the locus such that angle APB is a right angle.

P(x,y)
A(1,4)
B(-3,2)

Mpa * Mpb = -1

[(4-y)/(1-x)]*[(2-y)/(-3-x)] = -1

(4-y)(2-y) = -(1-x)(-3-x)
8 - 4y - 2y + y^2 = -(-3-x+3x+x^2)
8-6y + y^2 = -(-3+2x+x^2) = 3 - 2x - x^2

x^2 + 2x + y^2 - 6y = -5
(x+1)^2 + (y-3)^2 = -5 + 1 + 9 = 5

Circle with centre (-1,3) and radius of sqrt5.

 

[lolokay]

Help needed:
c) A(0,3) and B(-2,1), Given these points. Find the locus of the point P(x,y) so that APB is a right angle.

that one ^. it's locus

this one is just (x+1)² + (y-3)² = 5

 

[shaon0]

Re: 回复: Locus.

P(x,y)
A(1,4)
B(-3,2)

Mpa * Mpb = -1

[(4-y)/(1-x)]*[(2-y)/(-3-x)] -1

(4-y)(2-y) = -(1-x)(-3-x)
8 - 4y - 2y + y^2 = -(-3-x+3x+x^2)
8-6y + y^2 = -(-3+2x+x^2) = 3 - 2x - x^2

x^2 + 2x + y^2 - 6y = -5
(x+1)^2 + (y-3)^2 = -5 + 1 + 9 = 5

Circle with centre (-1,3) and radius of sqrt5.

thanks a lot.

 

[Affinity]

Re: 回复: Locus.

You should exclude points A and B from the circle though.. so it's a circle with 2 holes

 

[lolokay]

Re: 回复: Locus.

You should exclude points A and B from the circle though.. so it's a circle with 2 holes

argh, didn't think of that

 

[shaon0]

Re: 回复: Locus.

You should exclude points A and B from the circle though.. so it's a circle with 2 holes

Why?

 

[lolokay]

Re: 回复: Locus.

Why?

because if point P=A or B, a right angle is not formed

if you look at tommykins working, you'll see it involves dividing by 0

 

[tommykins]

回复: Re: 回复: Locus.

because if point P=A or B, a right angle is not formed

if you look at tommykins working, you'll see it involves dividing by 0

point's don't have an angle or magnitude (i think that's the word)