# Thread: Writing equation of locus

1. ## Writing equation of locus

A point P(x,y) moves such that it is equidistant from the points (-4,7) and (3,-2). What is the equation that represent this locus?

Thanks.

2. Originally Posted by shenton
A point P(x,y) moves such that it is equidistant from the points (-4,7) and (3,-2). What is the equation that represent this locus?

Thanks.

Hello, shenton,

the point P moves on the perpendicular bisector of the straight line from A(-4, 7) and B(3, -2).

1. Calculate the coordinates of the midpoint M of AB: M((-4+3)/2, (7-2)/2)
2. Calculate the slope of AB: (7-(-2))/(-4-3) = -9/7
3. The perpendicular direction is therefore: m = 7/9

Now you have a point and the slope of the line you are looking for. Use the point-slope-formula of a line:

(y - 5/2)/(x - (-1/2)) = 7/9 . Multiply by the denominator:

y - 5/2 = 7/9x + 7/18 Finally you get:

y = 7/9x + 26/9

EB

3. Hello, shenton,

I've attached a diagram of all the results of my calculations.

If you choose a point on the perpendicular bisector you'll see that it has the same distance to A and to B.

EB

4. Hello, shenton!

A point P(x,y) moves such that it is equidistant from the points A(-4,7) and B(3,-2).
What is the equation that represent this locus?
Hey, how about using the Distance Formula?

. . . . . . . - . - . . . . . ._______________
The distance PA is: . √(x + 4)² + (y - 7)²
. . . . . . . - . - . . . . . ._______________
The distance PB is: . √(x - 3)² + (y + 2)²

. . . . . . . . . . . . . _______________ . . - . ._______________
Since PA = PB: . √(x + 4)² + (y - 7)² . = . √(x - 3)² + (y + 2)²

Square both sides: . (x + 4)² + (y - 7)² . = . (x - 3)² + (y + 2)²

Expand: . x² + 8x + 16 + y² - 14y + 49 . = . x² - 6x + 9 + y² + 4y + 4

. . which simplifies to: . 14x - 18y + 52 .= .0 . . y .= .(7/9)x + 26/9