Higher June 2018 Paper 3 Q8
8
(a) Show that the lines \(\quad y = 3x + 7 \quad\) and \(\quad 2y - 6x = 8 \quad\) are parallel.
Do not use a graphical method. [3 marks]
(b) Is the point \((-5, -6)\) above, below or on the line \(y = 3x + 7\) ?
Tick one box.
- Above
- Below
- On the line
You must show your working.
Do not use a graphical method. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – Using gradients | ||
| Gradient of \(\;y = 3x + 7\;\) is 3 and \(\;y = 3x + 4\) and gradient of \(\;2y - 6x = 8\;\) is 3 or \(6 \div 2\) | B3 | May come from using points on line eg using (0, 7) and (1, 10) and \(\dfrac{10 - 7}{1 - 0} = 3\) or correct calculation for gradient from points on line \(2y - 6x = 8\) eg using (0, 4) and (1, 7) and \(\dfrac{7 - 4}{1 - 0} = 3\) B2 for \(y = 3x + 4\) and lines have same gradient or \(\;y = 3x + 4\) and gradient of \(\;2y - 6x = 8\;\) is 3 or \(6 \div 2\) or gradient of \(\;y = 3x + 7\;\) is 3 and \(\;y = 3x + 4\) B1 for gradient of \(\;y = 3x + 7\;\) is 3 or \(y = 3x + 4\) or gradient of \(\;2y - 6x = 8\;\) is 3 or \(6 \div 2\) |
| Alternative method 2 – Using coordinates and distances | ||
| Chooses a value for \(x\) and correctly evaluates the \(y\) value for both lines | M1 | eg (0, 7) and (0, 4) |
| Chooses a different value for \(x\) and correctly evaluates the \(y\) value for both lines | M1dep | eg (1, 10) and (1, 7) |
| States that \(y\) values are a constant distance apart so parallel | A1 | oe |
| Alternative method 3 – Using simultaneous equations | ||
| \(y = 3x + 4\) or \(\;y - 3x = 4\) or \(2y = 6x + 14\) or \(\;2y - 6x = 14\) | M1 | oe Equates coefficients in any form |
| Any attempt to eliminate both variables from their equations | M1dep | |
| States simultaneous equations have no (real) solution and concludes parallel | A1 | |
Additional guidance
| To award A mark on Alternative method 2, the working must be seen | |
| \(y = 3x + 4\) and lines have gradient of \(3x\) | B2 |
| \(y = 3x + 4\) and \(3x\) identified in both equations | B2 |
| Both lines have gradient \(3x\) | B1 |
| \(y = 3x + 7\), gradient 3 and \(y = 3x + 8\), gradient 3 (error in rearrangement) | B1 |
| \(y = 3x + 8\), gradient 3 (error in rearrangement) | B0 |
| Parallel as both have same gradient | B0 |
| \(2(3x + 7) - 6x = 8\) \(6x + 14 - 6x = 8\) \(14 = 8\) | M1 M1 |
| \(y = 3x + 7\) and \(y = \dfrac{8 + 6x}{2}\) are equated coefficients, Alternative method 3 | M1 |
| Answer | Mark | Comments |
|---|---|---|
| \(3 \times -5 + 7\) or \(-15 + 7\) or \(-8\) or \((-5, -8)\) or \((-6 - 7) \div 3\) or \(-4.33\)… or \(\;y = 3x + 9\) | M1 | Use a point on \(y = 3x + 7\) with \((-5, -6)\) to compare gradient to 3 eg Gradient from \((-5, -6)\) to \((0, 7)\) is 2.6 |
| Above and \(-8\) or Above and \(-4.33\) or Above and \(\;y = 3x + 9\) | A1 | oe Above and eg Gradient from \((-5, -6)\) to \((0, 7)\) is 2.6 |
Additional guidance
| Do not ignore incorrect statements eg \(-6\) is less than \(-8\) so above | M1A0 |
| \((0, 7)\), \((-1, 4)\), \((-2, 1)\), \((-3, -2)\), \((-4, -5)\), \((-5, -8)\) and ticks below | M1A0 |