# Thread: locus of points equidistand to both a point and straight line(horizontal)

1. ## locus of points equidistand to both a point and straight line(horizontal)

P(x;y). Is a point which is equidistant to both point D(2;-5) and straight line y=1
Determine the equation of locuss of P(x;y)
I drew horizontal y=1 and plot D(2;-5). Am not sure how to draw locus of points which satisfies given conditions?

2. ## Re: locus of points equidistand to both a point and straight line(horizontal)

Hint: A locus of points equidistant from a point and a line is a parabola. A quadratic equation models a parabola.

3. ## Re: locus of points equidistand to both a point and straight line(horizontal)

I need to sketch it before goin on to find the equation.

4. ## Re: locus of points equidistand to both a point and straight line(horizontal)

Hey Bonganitedd.

The distance between two points is d^2 = x^2 + y^2 where x is the x-distance and y is the y-distance.

For a circle this is (x-a)^2 + (y-b)^2 = r^2 and for the line you have y = 0 or y = 2 (since in euclidean space, you have parallel lines for constant distance).

So now you have three equations and you must solve for these.

5. ## Re: locus of points equidistand to both a point and straight line(horizontal)

@ Chiro I think u r complicating the problem.I did try using definitions of loci,after having (1) equation of circle => (x-2)^2 +(y+5)^2 = 29 (2) y=0 and (3) y=2. How do I work the equation because the equation must in the form of y = ax^2 +bx+c. Its more of manipulating than going with definitions. If I get my sketch right I can equate the distances from the line y=1 to P(x;y) and from P(x;y) to D(2;-5)

6. ## Re: locus of points equidistand to both a point and straight line(horizontal)

The distance from the point (2, -5) to the point (x, y) is $\displaystyle \sqrt{(x- 2)^2+ (y+ 5)^2}$. The distance from the line y= 1 to the point (x, y) is y- 1. Set those equal and simplify.

7. ## Re: locus of points equidistand to both a point and straight line(horizontal)

Whatever works for you Bonganitedd: there are many ways to solve a single problem in mathematics.

8. ## Re: locus of points equidistand to both a point and straight line(horizontal)

Lot many suggestions have been given i would like to just add that we have an expression for finding distance of a point ( p,q) from a line ax+by+c=0 given by
[ |ap+bq+c|]/[ sqrt ( a^2 +b^2)] Use this and the other expression given for the distance between two points and you will get the locus.

9. ## Re: locus of points equidistand to both a point and straight line(horizontal)

This another attempt of mine after many suggestions. I set a third point along the line y=1 I call this point Q(x;1) then distance from PQ =PD i.e PQ^2 =PD^2
d^2 =(x2-x1)^2 +(y2-y1)^2 substituting we have
(x-x)^2 + (y-1)^2 = (2-x)^2 + (-5-y)^2
y^2 -2y +1=x^2 -4x +4 +y^2 +10y +25
12y = -x^2 +4x -28

10. ## Re: locus of points equidistand to both a point and straight line(horizontal)

The line is called directrice, and the fixed point is called focus. Prove It told you what is the name of the curve. Are you still confused?

11. ## Re: locus of points equidistand to both a point and straight line(horizontal)

@ Vito I get it now.