# Math Help - Conics, Lines questions

1. ## Conics, Lines questions

1)Find the equation(s) of the locus of point P that is a constant distance of 4 units from the locus defined by y=2x+1.

No idea how to go about doing this

2) Find the equation of the locus of a point P that is equidistant from A(-4,4 and B(4,2).

same for this question
plz help me.

2. Originally Posted by Solid8Snake
1)Find the equation(s) of the locus of point P that is a constant distance of 4 units from the locus defined by y=2x+1.

No idea how to go about doing this

2) Find the equation of the locus of a point P that is equidistant from A(-4,4 and B(4,2).

same for this question
plz help me.
to #1:

The locus consists of 2 parallels to the given line with a distance of 4.

The equation of the given line is: $-2x+y-1=0$

Now use the formula to calculate the distance of a point P(s, t) to a straight line:

$d = \dfrac{As + Bt + C}{\sqrt{A^2+B^2}}$

The distance of point P(x, y) to the given line is therefore:

$d = 4 = \dfrac{-2x + y - 1}{\sqrt{2^2+1^2}}~\implies~y = -2x+1 \pm 4\sqrt{5}$

to #2

The locus is the perpendicular bisector of the line segment AB.

Let P have the coordinates (x, y). Calculate the distance AP and the distance BP. Both must be equal. Solve for y.

3. I don't quite understand how you +-4root5 on the either side, care to explain?

4. Originally Posted by Solid8Snake
I don't quite understand how you +-4root5 on the either side, care to explain?
The distance of point P(x, y) to the given line is therefore:

$4 = \dfrac{-2x + y - 1}{\sqrt{2^2+1^2}}~\implies~4 = \dfrac{-2x + y - 1}{\sqrt{5}}$

Now multiply both sides of the equation by $\sqrt{5}$ and solve for y.
You'll get the result I've given in my previous post.

5. Thank you very much!