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.
If B is a point, how can a line be perpendicular to it?
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
To derive an equation specifying the locus of point P, we need to use the concept of slopes and perpendicular lines.
Let's start by finding the slope of the line joining points A(9, -5) and B(4, 5):
Slope of AB = (change in y) / (change in x)
= (5 - (-5)) / (4 - 9)
= 10 / (-5)
= -2
Since the line joining P to A is perpendicular to line AB, the product of their slopes must be -1:
Slope of line AB * Slope of line PA = -1
Let's assume the coordinates of point P are (x, y). The slope of line PA would then be:
Slope of line PA = (y - (-5)) / (x - 9)
= (y + 5) / (x - 9)
We can now set up the equation using the product of slopes:
(y + 5) / (x - 9) * (-2) = -1
Simplifying the equation by cross-multiplying:
-2(y + 5) = (x - 9)
Expanding and rearranging the equation:
-2y - 10 = x - 9
-2y = x + 1
Finally, we can rearrange the equation to have y as the subject:
y = (1/2)x - 1/2
Hence, the equation specifying the locus of point P is y = (1/2)x - 1/2.