a) find the equation of AB, the perpindicular bisector of the line joining the points P(-3,1) and Q(1,9)

(line AB intersects the point at midway of PQ)

b) C is the centre of a circle passing through P and Q. given that QC is parralel to the yaxis, determine theequation of the circle.

(c) the tangents at P and Q intersect at T

(1) find the equation of the tangent at Q

(2) the coordinates at C

i know that it is difficult to answer this question without a diagram so i am leaving my email address here so anywone that is kind enough to help someone who is on the verge of sitting his exams and requiring urgent help..if you can sent me a short message just saying that you are willing to help me then i will definietly sent you a copy of my question..... just remember that i will always be grateful for helping me

my email is adilahmed1 at the rate of btinternet com

a) To find the equation of the perpendicular bisector of the line joining points P(-3,1) and Q(1,9), we can follow these steps:

1. Find the midpoint of PQ:
The midpoint of a line segment can be found by averaging the x-coordinates and averaging the y-coordinates of its endpoints.
Midpoint M = [(x1 + x2)/2, (y1 + y2)/2] = [(-3+1)/2, (1+9)/2] = [-1, 5]

2. Find the slope of PQ:
The slope of a line passing through two points (x1, y1) and (x2, y2) is given by:
m = (y2 - y1) / (x2 - x1)
m = (9 - 1) / (1 - (-3)) = 8 / 4 = 2

3. The slope of the perpendicular bisector is the negative reciprocal of the slope of PQ.
Perpendicular bisector slope = -1 / m = -1 / 2 = -0.5

4. Use the point-slope form to find the equation of the perpendicular bisector:
The point-slope form of a line with slope m passing through point (x1, y1) is:
y - y1 = m(x - x1)

Using the midpoint M (-1, 5):
y - 5 = -0.5(x - (-1))
y - 5 = -0.5(x + 1)
y - 5 = -0.5x - 0.5
y = -0.5x + 4.5

The equation of AB, the perpendicular bisector of PQ, is y = -0.5x + 4.5.

b) To determine the equation of the circle passing through P and Q, let's follow these steps:

1. Find the midpoint of PQ (already calculated in part a) as M(-1, 5).

2. Find the radius of the circle, which is the distance from the center to either P or Q.
radius = distance from M to either P or Q = distance formula
radius = sqrt((x2 - x1)^2 + (y2 - y1)^2)
radius = sqrt((-1 - (-3))^2 + (5 - 1)^2) = sqrt(4^2 + 4^2) = sqrt(32) = 4sqrt(2)

3. Determine the equation of the circle with center C and radius 4sqrt(2).
The equation of a circle with center (h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2

Since QC is parallel to the y-axis, we know that the x-coordinate of C is the same as the x-coordinate of Q.
Center C = (1, k)

Using the radius 4sqrt(2):
(x - 1)^2 + (y - k)^2 = (4sqrt(2))^2
(x - 1)^2 + (y - k)^2 = 32

The equation of the circle is (x - 1)^2 + (y - k)^2 = 32.

c) For this part, we need more information to answer the questions. Could you please provide additional details or the complete question?