(i) Gradient of BC = -2

Since BC is perpendicular to CD, gradient of CD = 1/2

Equation of CD, using gradient = 1/2 and point D (0, 3):

                   y - 3 = 1/2 x

                      2y = x + 6     ....(1)

 

(ii) y = -2x + 13  ...(2)

Solving (1) and (2) simultaneously,

             x = 4, y = 5

The coordinates of C are (4, 5)

 

(iii) Midpoint of BD = (5/2, 3)

 

Let A be (x, y)

 

([x+4]/2, [y+5]/2) = (5/2, 3)

---> x + 4 = 5 , x = 1

---> y + 5 = 6, y = 1

The coordinates of A are (1, 1)

 

 

(i) Coordinates of M, midpoint of OB = (5/2, 3/2)

 

(ii) Gradient of OB = 3/5

     Gradient of AC, perpendicular to OB = -5/3

     Equation of AC:      y - 3/2 = -5/3 (x - 5/2)

                               3y + 5x = 17

 

(iii) At A (7, a),

            3a + 5(7) = 17

                       a =  -6

 

(iv) At C (c, c+3),

            3 (c + 3) + 5c = 17

                              c = 1

The coordinates of C are (1, 4)

 

(v) Area

 

(vi)