Points A (-2, 3) and B (3, -2), lie on the circle x2+y2=13.
Determine the equation of the perpendicular bisector of AB.
To find the equation of the perpendicular bisector of AB, we need to find the midpoint of AB and the slope of AB.
First, let's find the midpoint of AB:
The x-coordinate of the midpoint, M, is the average of the x-coordinates of A and B:
x-coordinate of M = (x-coordinate of A + x-coordinate of B) / 2 = (-2 + 3) / 2 = 1/2.
The y-coordinate of the midpoint, M, is the average of the y-coordinates of A and B:
y-coordinate of M = (y-coordinate of A + y-coordinate of B) / 2 = (3 + (-2)) / 2 = 1/2.
Therefore, the midpoint of AB is M(1/2, 1/2).
Next, let's find the slope of AB:
The slope, m, of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:
m = (y2 - y1) / (x2 - x1).
Using the coordinates of A and B, we have:
m = (-2 - 3) / (3 - (-2)) = -5 / 5 = -1.
Therefore, the slope of AB is -1.
Since the perpendicular bisector of a line has a slope that is the negative reciprocal of the original line's slope, the slope of the perpendicular bisector of AB is 1.
Finally, we can use the midpoint M(1/2, 1/2) and the slope 1 to find the equation of the perpendicular bisector using the point-slope form of a line:
y - y1 = m(x - x1).
Using M(1/2, 1/2) and m = 1, we have:
y - 1/2 = 1(x - 1/2).
Simplifying, we find:
y - 1/2 = x - 1/2.
Adding 1/2 to both sides, we have:
y = x.
Therefore, the equation of the perpendicular bisector of AB is y = x.
Show that the perpendicular bisector of AB passes through the centre of the circle.
To show that the perpendicular bisector of AB passes through the center of the circle, we need to show that the coordinates of the midpoint of AB are the same as the coordinates of the center of the circle.
The equation of the circle is x^2 + y^2 = 13.
We found earlier that the midpoint of AB is M(1/2, 1/2).
To show that M is the center of the circle, we need to show that M satisfies the equation of the circle.
Substituting x = 1/2 and y = 1/2 into the equation of the circle, we have:
(1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2.
Since 1/2 is equal to 13, we can conclude that the midpoint M(1/2, 1/2) lies on the circle x^2 + y^2 = 13.
Therefore, the perpendicular bisector of AB passes through the center of the circle.