Find the orthocentre of the vertices QRS. Q(2,6), R (-3,1) and S (6,2).

To find the orthocentre of a triangle, you need to follow these steps:

1. Find the slopes of the lines containing the sides of the triangle QRS.
2. Determine the slopes of the lines perpendicular to these sides.
3. Find the equations of the perpendicular bisectors for any two sides.
4. Solve the equations of the bisectors to find their point of intersection, which is the orthocentre.

Let's go through these steps to find the orthocentre of triangle QRS:

1. Finding slopes of sides:
The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:
slope = (y2 - y1) / (x2 - x1)

Slope of line QR:
slope_QR = (1 - 6) / (-3 - 2) = -5 / -5 = 1

Slope of line RS:
slope_RS = (2 - 1) / (6 - (-3)) = 1 / 9

Slope of line SQ:
slope_SQ = (6 - 2) / (2 - 6) = 4 / (-4) = -1

2. Finding slopes of perpendicular lines:
The slopes of the perpendicular lines will be the negatives of the reciprocals of the slopes of the sides.

Slope of perpendicular to QR:
slope_perpendicular_QR = -1 / slope_QR = -1 / 1 = -1

Slope of perpendicular to RS:
slope_perpendicular_RS = -1 / slope_RS = -1 / (1 / 9) = -9

Slope of perpendicular to SQ:
slope_perpendicular_SQ = -1 / slope_SQ = -1 / (-1) = 1

3. Finding equations of perpendicular bisectors:
The perpendicular bisector of a line segment is the line passing through the midpoint of the segment and perpendicular to it.
We need to find two perpendicular bisectors.

a) Perpendicular bisector of QR:
- Finding midpoint M of QR:
Mx = (x1 + x2) / 2
My = (y1 + y2) / 2
Mx = (2 - 3) / 2 = -1 / 2
My = (6 + 1) / 2 = 7 / 2

- Finding equation of the perpendicular bisector passing through M:
Using point-slope form of a line: y - y1 = m(x - x1)
y - (7/2) = -1(x - (-1/2))
y - (7/2) = -x - 1/2
y = -x - 1/2 + 7/2
y = -x + 6/2
y = -x + 3

b) Perpendicular Bisector of RS:
- Finding midpoint N of RS:
Nx = (x1 + x2) / 2
Ny = (y1 + y2) / 2
Nx = (-3 + 6) / 2 = 3/2
Ny = (1 + 2) / 2 = 3/2

- Finding equation of the perpendicular bisector passing through N:
Using point-slope form of a line: y - y1 = m(x - x1)
y - (3/2) = -9(x - (3/2))
y - (3/2) = -9x + 27/2
y = -9x + 27/2 + 3/2
y = -9x + 30/2
y = -9x + 15

4. Solving the equations of the bisectors to find the orthocentre:
To find the orthocentre, we need to find the point of intersection of the two perpendicular bisectors.

Equating the equations of the bisectors:
-x + 3 = -9x + 15

Solving for x:
8x = 12
x = 12 / 8
x = 3 / 2

Substituting x back into either equation (let's use the first one):
y = -x + 3
y = -(3/2) + 3
y = 3/2

Therefore, the orthocentre of triangle QRS is at the point (3/2, 3/2).