(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) \(2 \Delta n = 5 \Delta 7\).
(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.
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