The point A has coordinates (-1, 6) and the point B has coordinates (7, 2). i Find the equation of the perpendicular bisector of AB, giving your answer in the form y = mx + c.

ii A point C on the perpendicular bisector has coordinates (p, q). The distance OC is 2 units, where O is

the origin. Write down two equations involving p and q and hence find the coordinates of the possible

positions of C.

a) y = 2x -2

b) Equation: q = 2p - 2  (1)

p² + q² = 4   (2)

The coordinates of C may be (0, -2) or (6/5, 8/5)

Step-by-step explanation:

a) Step 1: Find the coordinates of midpoint, P of AB

x-co-ordinate of P = (x₁ + x₂)/2 = (-1 + 7)/2 = 3

y- co-ordinate of P = (y₁ + y₂)/2 = (6 + 2)/2 = 4

Step 2: Determine the gradient, m₂ of perpendicular bisector of AB

For perpendicular lines, m₁ * m₂ = -1 where m₁ is slope of AB

m₁ = (2 - 6)/(7 - -1) = -4/8 = -1/2

-1/2 * m₂ = -1

m₂ = 2

Step 3: Deriving the equation of the perpendicular bisector in the form y = mx + c

y - y₁ = m(x - x₁); substituting the values of the midpoints of the line

y - 4 = 2(x - 3)

y - 4 = 2x - 6

y = 2x -2

b) Step 1: Derive the values of the p and using the formula of distance between two points:

y = 2x - 2

therefore q = 2p - 2  (1)

OC = √(x₂ -x₁)² + (y₂ - y₁)²

2 = √(p - 0)² + ( q - 0)²

2 = √p² + q²

squaring both sides

p² + q² = 4   (2)

Substituting (1) in (2)

p² + (2p - 2)² = 4

p² + 4p² - 8p + 4 = 4

5p² - 8p = 0

p(5p - 8) = 0

Therefore p = 0 or 5p - 8 = 0

p = 0 or p = 8/5

Substituting p = 0 in (1)

q = -2

substituting p = 8/5 in (1)

q = 2(8/5) - 2

q = 16/5 - 2

q = 6/5

Hence the coordinates of C may be (0, -2) or (6/5, 8/5)

  y= -3x is the required function.

step-by-step explanation:

we have been given a table:

x             y

0             0

1             -3

3             -9

-2             6

we can see that the values of y is -3 times the value of x

hence, the function would be: y= -3x.

put the values of x will give the same y as in the table that means function is satisfying the given table.

two data sets with the same median will not have the same varaibility

step-by-step explanation: