(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 ...
Question 1 Report
(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) \(2 \Delta n = 5 \Delta 7\).
Answer Details
(a) Equation of the line
\(2x+y=6\Rightarrow y=-2x+6\), so its gradient is \(-2\). A parallel line has the same gradient \(m=-2\) and passes through \((2,-3)\):
\[y-(-3)=-2(x-2)\Rightarrow y+3=-2x+4\Rightarrow y=-2x+1\quad\text{i.e. }2x+y=1.\]
(b)(i) With \(x\,\Delta\,y=(x+y+xy)\bmod 8\) on \(T=\{2,3,5,7\}\):
| \(\Delta\) | 2 | 3 | 5 | 7 |
|---|---|---|---|---|
| 2 | 0 | 3 | 1 | 7 |
| 3 | 3 | 7 | 7 | 7 |
| 5 | 1 | 7 | 3 | 7 |
| 7 | 7 | 7 | 7 | 7 |
(e.g. \(2\Delta5=(2+5+10)\bmod 8=17\bmod 8=1\).)
(ii)(a) \(5\Delta7=7\), so \(2\Delta(5\Delta7)=2\Delta7=7\).
(ii)(b) \(5\Delta7=7\), so we need \(2\Delta n=7\). From the \(2\)-row, \(2\Delta7=7\), hence \(n=7\).
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