Locus of a Point: Find Equation for Equidistant Point P from (3,-1)

[zebra1707]

1. Find the equation of the locus of a point P which is equidistant from the y-axis and the point (3,-1)


2. I think that I need to use

SqRoot (x-x)sq + (y-y)sq = x
SqRoot (x-3)sq + (y+1)sq = x

Expanding is where I get stuck



Cheers

 

[statdad]

If $$(a,b)$$ is the point, you need work with

Distance from $$(a,b)$$ to $$(3,-1)$$ = distance from $$(a,b)$$ to the $$y-$$ axis. Part of this equation is

$$\sqrt{(a-0)^2 + (b-b)^2} = \sqrt{a^2} = |a|$$

What is the other part? remember your solution will be an equation, not a single number.

 

[zebra1707]

Hi there
There is no other part to this question. I think the equation should be y=x Sq + 3

Cheers (very confused).