sry bout that , but i got aactual problem (1 Viewer)

[wolf7]

find the equation of the locus of point p(x,y) that moves so that it is equidistant from the x-axis and the y-axis

also

find the equation of the locus of a point P that moves so that PA is twice the distance of PB where A = (0,3) and B = (3,-1)

 

[Trev]

Equidistant from (x,0) and (0,y)
√[(x-x)²+(y-0)²] = √[(x-0)²+(y-y)²]
y²=x²
y=+/-x

 

[insert-username]

find the equation of the locus of a point P that moves so that PA is twice the distance of PB where A = (0,3) and B = (3,-1)

Let P = (x,y).

PA = 2PB

Therefore (PA)² = 4(PB)²

[x² + (y-3)²] = 4[(x-3)² + (y+1)²]

x² + y² - 6y + 9 = 4(x² - 6x + 9 + y² + 2y + 1)

x² + y² - 6y + 9 = 4x² - 24x + 36 + 4y² + 8y + 4

0 = 3x² - 24x + 3y² + 14y +31

0 = (√3x - 4√3)² + ....


I got to there before I got bogged totally. Can a 3x² be factorised into a perfect square, or have I gone wrong somewhere?


I_F

 

[Trev]

3x² - 24x + 3y² + 14y +31 = 0
x² - 8x + y² + (14/3)y + 31/3 = 0
(x-4)²-16 + (y+(7/3))²-(7/3)² + 31/3 = 0
etc.

 

[Riviet]

For the first one, you could also try the geometrical method, which sometimes is much quicker than the algebraic method. The mathematically correct answer is really y=+/- |x| from y²=x²