find equation of the locus of point P if?

Question

if the line joining P to A (-2,1) is perpendicular to the line that joins P to B(6,5).

Answer 1

The locus is a circle with the line segment AB as its diameter!

Remember from 9th grade geometry that if you start with a circle, pick a diameter, pick an arbitrary point and connect it to the endpoints of the diameter, the angle formed is always a right angle.

Answer 2

Let the coordinates of P be x, y. Then the slope of the line joining P and A would be (y-1)/(x+2) and the slope of the line joining P and B will be (y-5)/(x-6). According to the problem, these lines are perpendicular. So, the product of their slopes will be -1. Thus, (y-1)(y-5)=-(x-6)(x+2)
or, y^2+x^2-6y-4x-7=0
This is the equation of a circle with center at (2,3) and radius 2*sqrt(5) as we can see from (y-3)^2+(x-2)^2=(2sqrt5)^2

Answer 3

It just so happens that a circle with diameter AB fits the requirements (except, of course, at A and B).
(x - 2)^2 + (y - 3)^2 = 20

You can also get there by the Theoem of Pythagorus:
(x - 6)^2 + (y - 5)^2 + (x - (-2))^2 + (y - 1)^2 = (6 - (-2)^2 + (5 - 1)^2
(x^2 - 12x + 36) + (y^2 - 10y + 25) + (x^2 + 4x + 4) + (y^2 - 2y + 1) = (6 + 2)^2 + 4^2
2x^2 - 8x + 40 + 2y^2 - 12y + 26 = 64 + 16
x^2 - 4x + 20 + y^2 - 6y + 13 = 40
x^2 - 4x + y^2 - 6y = 7
x^2 - 4x + 4 + y^2 - 6y + 9 = 7 + 4 + 9
(x - 2)^2 + (y - 3)^2 = 20

Answer 4

Sketch the situation and put a point P (x, y) on it. Join P to A and B. Find the gradient of PA and of PB by the method of
(difference in y coordinates)/(difference in x coordinates).
Since the lines PA and PB are perpendicular the two gradients must have a product of -1. When you do this and rearrange your equation you will find that it comes out to be the equation of a circle going through A and B.

Answer 5

i could like to grant this a crack, yet regrettably, i'm late because it fairly is. the way i could attitude it may be to construct a parabola employing the concentration-directorix shape. you ought to use a vector equation for the directorix. replace in the coordinates of the factors to get some circumstances for the variables in the parabola's shape. with any luck, those circumstances will spit out an equation for the locus.

Answer 6

A sketch suggests that the locus is just two points.

The line between A and B is the diagonal of a rectangle. The other two corners is the locus.