June 2017 Paper 2 Q14
14 The line \(\quad y = ax + b \quad\) is perpendicular to the line \(\quad x + 4y = 74\)
The lines intersect at the point where \(x = 2\)
Not drawn accurately
Work out the values of \(a\) and \(b\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(-\dfrac{1}{4}\) or \(-0.25\) | B1 | gradient of \(x + 4y = 74\) Do not allow embedded May be implied |
| (gradient =) \(\dfrac{-1}{\text{their } -\frac{1}{4}}\) or 4 | M1 | ft their \(-\dfrac{1}{4}\) Only ft a non-zero numerical value Implied by \(y = 4x + b\) or \(a = 4\) (B1M1) |
| (\(y\) =) \(\dfrac{74 - 2}{4}\) or \(\dfrac{72}{4}\) or 18 | M1 | oe May be seen on diagram |
| their \(18 =\) their \(4 \times 2 + b\) or \(y -\) their \(18 =\) their \(4(x - 2)\) | M1dep | oe dep on M2 |
| \(b = 10\) | A1ft | ft \(18 -\) their \(4 \times 2\) if B0M3 |
| Alternative method 2 | ||
| \(-\dfrac{1}{4}\) or \(-0.25\) | B1 | gradient of \(x + 4y = 74\) Do not allow embedded May be implied |
| (gradient =) \(\dfrac{-1}{\text{their } -\frac{1}{4}}\) or 4 | M1 | ft their \(-\dfrac{1}{4}\) Only ft a non-zero numerical value Implied by \(y = 4x + b\) or \(a = 4\) (B1M1) |
| Correct method for elimination of \(y\) from \(x + 4y = 74\) and \(y =\) their \(4x + b\) | M1dep | eg \(x + 4(4x + b) = 74\) or \(17x + 4b = 74\) |
| Substitutes \(x = 2\) into their equation | M1dep | eg \(34 + 4b = 74\) |
| \(b = 10\) | A1ft | ft \(18 -\) their \(4 \times 2\) if B0M3 |
| Alternative method 3 | ||
| \(-\dfrac{1}{4}\) or \(-0.25\) | B1 | gradient of \(x + 4y = 74\) Do not allow embedded May be implied |
| (gradient =) \(\dfrac{-1}{\text{their } -\frac{1}{4}}\) or 4 | M1 | ft their \(-\dfrac{1}{4}\) Only ft a non-zero numerical value Implied by \(y = 4x + b\) or \(a = 4\) (B1M1) |
| (\(y\) =) \(\dfrac{74 - 2}{4}\) or \(\dfrac{72}{4}\) or 18 | M1 | oe May be seen on diagram |
| Correct method for elimination of \(x\) from \(x + 4y = 74\) and \(y =\) their \(4x + b\) and substitutes \(y =\) their 18 | M1dep | eg \(y = 4(74 - 4y) + b\) or \(17y = 296 + b\) and \(306 = 296 + b\) dep on M2 |
| \(b = 10\) | A1ft | ft \(18 -\) their \(4 \times 2\) if B0M3 |
Additional guidance
\(y = 4x + 10\) will gain full marks unless contradicted
If an error is made in the constant term when rearranging \(x + 4y = 74\)
the B1 can still be awarded for gradient \(= -\dfrac{1}{4}\)
eg \(y = -\dfrac{1}{4}x + 19\) and gradient \(= -\dfrac{1}{4}\) is B1
(all other marks are possible)
In alt 1 and alt 3 the mark for \(y = 18\) will sometimes be the only mark awarded