help wit locus q plz? (1 Viewer)

[Trev]

hey can someone plz show me how to do this q plz - (if u can b bothered can u do the second one too plz)

Q1) 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

Q2) Find the equation of the locus of point P(x,y) that moves so that the ratio of PA to PB is 3:2 where A = (-6,5) and B= (3,-1)


Answers (so u can check if u did it right)
A1)y=+x, -x A2)5x² - 102x + 5y² + 58y - 154 = 0

thanx cya

 

[CM_Tutor]

(1) The distance from P(x, y) to the x-axis is |y|, and the distance to the y-axis is |x|. These distances are the same, so the locus is |y| = |x|, or y = +/- x.

(2) PA² = (x + 6)² + (y - 5)²
PB² = (x - 3)² + (y + 1)²

Now PA / PB = 3 / 2 (given)
So, 2PA = 3PB
4PA² = 9PB²
4[(x + 6)² + (y - 5)²] = 9[(x - 3)² + (y + 1)²]
4[x² + 12x + 36 + y² - 10y + 25] = 9[x² - 6x + 9 + y² + 2y + 1]
4x² + 48x + 244 + 4y² - 40y = 9x² - 54x + 90 + 9y² + 18y

So, 5x² - 102x + 5y² +58y - 154 = 0, as required.