Locus Question (1 Viewer)

[AwesomeJelly]

Find the locus, if the distance of a point from y= -5 is three quarters of its distance from x=2

how do u work this out?

 

[Drongoski]

Find the locus, if the distance of a point from y= -5 is three quarters of its distance from x=2

how do u work this out?

Let d1 = distance from y = -5 and d2 = distance from x=5

Let P(x,y) be a typical point satisfying conditions of locus

.: d1 = |y - (-5)|

d2 = |x - 2|

Then solve the eqn: |y - (-5)| = 0.75|x-2|

 

[largarithmic]

Let P(x,y) be a point in the plane.

y = -5 -> A(x, -5)
x = 2 -> B(2, y)

3PA/4 = PB
3PA = 4PB
3(sqrt((y+5)^2)) = 4(sqrt(x-2)^2)
3(y+5) = 4(x-2)
3y + 15 = 4x-8
4x - 3y - 27 = 0

You're missing a solution that you'd obtain from the 3(y+5) = -4(x-2) case (i.e. 4x + 3y + 7 = 0). Also you got the arithmetic wrong, -8-15= -23

 

[nightweaver066]

You're missing a solution that you'd obtain from the 3(y+5) = -4(x-2) case (i.e. 4x + 3y + 7 = 0). Also you got the arithmetic wrong, -8-15= -23

Thanks for the corrections, haven't touched locus for a while now. Silly mistakes always get the better of me..

 

[Drongoski]

I got the locus as the 2 straight lines:

and

or equivalently:

3x-4y-26 = 0 and 3x+4y+14 = 0

 

[AwesomeJelly]

oh i got it xD
x=2 so let PA be (2,y)
y=-5 so let PB be (x,-5)

|PB|=3/4|PA|
PB=y-(-5)
=y+5
PA=x-2

y+5=3/4(x-2)
4y+20=3x-6
3x-4y-26=0

PB=-(y+5)
PA=x-2
-y-5=3/4(x-2)
-4y-20=3x-6
3x+4y+14=0

Thanks for the help