(a) Find the equation of a straight line which passes through the point (2, -3) and is parallel to the line (2x + y = 6).
(b) The operation (Delta) is defined on the set T = {2, 3, 5, 7} by (x Delta y = (x + y + xy) mod 8).
(i) Construct modulo 8 table for the operation (Delta) on the set T.
(ii) Use the the table to find: (a) (2 Delta (5 Delta 7)) ; (b) (2 Delta n = 5 Delta 7).
Explanation
(a) First, change the given linear equation to the gradient- intercept form:
(2x + y = 6 implies y = 6 – 2x)
Thus, (m_{1} = -2, c = 6). Thus, for a line to be parallel, the gradient (m_{2} = -2) (condition for parallelism).
Using (y – y_{1} = m(x – x_{1})), we have
(y – (-3) = -2(x – 2))
(y + 3 = – 2x + 4)
(2x + y = 4 – 3 = 1)
Thus, the line passing through (2, -3) parallel to 2x + y = 6 is 2x + y = 1.
(b) (i)
| (Delta) | 2 | 3 | 5 | 7 |
| 2 | 0 | 3 | 1 | 7 |
| 3 | 3 | 7 | 7 | 7 |
| 5 | 1 | 7 | 3 | 7 |
| 7 | 7 | 7 | 7 | 7 |
(ii) (a) (2 Delta (5 Delta 7) = 2 Delta 7 = 7)
(b) (2 Delta n = 5 Delta 7)
(2 Delta n = 7)
By comparison, n = 7.