Asked by Gemma
A point P(x,y) moves so that the line joining P to A(9,-5) is perpendicular to B(4,5). Derive an equation specifying the locus of P.
Answers
Answered by
Steve
If B is a point, how can a line be perpendicular to it?
Answered by
Reiny
You must have meant:
"... so that the line joining P to A(9,-5) is perpendicular to P joining B(4,5)
Recall this property of a circle:
If any point P on the circle is joined to the endpoint A and B of a diameter, the AP is perpendicular to BP
Thus AB must be a diameter and its midpoint is the centre of the circle
midpoint of AB = ((9+4)/2 , (-5+5)/2)
= ( 13/2, 0 )
so the equation is a circle of the form
(x - 13/2)^2 + y^2 = r^2
plut in (4,5), since it lies on the circle
(4-13/2)^2 + 25 = r^2
r^2 = 125/4
(x - 13/2)^2 + y^2 = 125/4
"... so that the line joining P to A(9,-5) is perpendicular to P joining B(4,5)
Recall this property of a circle:
If any point P on the circle is joined to the endpoint A and B of a diameter, the AP is perpendicular to BP
Thus AB must be a diameter and its midpoint is the centre of the circle
midpoint of AB = ((9+4)/2 , (-5+5)/2)
= ( 13/2, 0 )
so the equation is a circle of the form
(x - 13/2)^2 + y^2 = r^2
plut in (4,5), since it lies on the circle
(4-13/2)^2 + 25 = r^2
r^2 = 125/4
(x - 13/2)^2 + y^2 = 125/4
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