June 2023 Paper 2 Q4
4 A line passes through \(P\,(1, k)\) and \(Q\,(r, 6)\) where \(k\) and \(r\) are constants.
The midpoint of \(PQ\) has \(x\)-coordinate 5
The gradient of the line is 2
Work out the value of \(k\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Works out \(r\) and uses it to work out gradient expression for \(PQ\) | ||
| \(\dfrac{1 + r}{2} = 5\) or (\(r =\)) \(5 \times 2 - 1\) or (\(r =\)) \(1 + 4 + 4\) or (\(r =\)) 9 | M1 | oe equation or calculation eg \(\dfrac{r - 1}{2} + 1 = 5\) or (\(r =\)) \(5 + 5 - 1\) may be seen on a sketch implied by \(Q\,(9, 6)\) implied by correct gradient expression in \(k\) |
| \(\dfrac{6 - k}{\text{their } 9 - 1}\) or \(\dfrac{6 - \frac{6 + k}{2}}{\text{their } 9 - 5}\) | M1 | oe expression their 9 must be a value their 9 cannot be 1 or 5 |
| \(\dfrac{6 - k}{\text{their } 9 - 1} = 2\) or \(\dfrac{6 - \frac{6 + k}{2}}{\text{their } 9 - 5} = 2\) | M1dep | oe equation eg \(\dfrac{k - 6}{1 - \text{their } 9} = 2\) or \(6 - k = 16\) or \(\dfrac{6 + k}{2} = -8 + 6\) dep on 2nd M1 |
| \(-10\) | A1 | |
| Alternative method 2 Works out \(r\) and uses it to work out equation of line \(PQ\) | ||
| \(\dfrac{1 + r}{2} = 5\) or (\(r =\)) \(5 \times 2 - 1\) or (\(r =\)) \(1 + 4 + 4\) or (\(r =\)) 9 | M1 | oe equation or calculation eg \(\dfrac{r - 1}{2} + 1 = 5\) or (\(r =\)) \(5 + 5 - 1\) may be seen on a sketch implied by \(Q\,(9, 6)\) |
| \(6 = 2 \times \text{their } 9 + c\) or \(c = -12\) or \(y = 2x - 12\) | M1 | oe their 9 must be a value their 9 cannot be 1 or 5 |
| \(2 \times 1 + 6 - 2 \times \text{their } 9\) | M1dep | oe eg \(2 - 12\) dep on 2nd M1 |
| \(-10\) | A1 | |
| Alternative method 3 Works out \(r\) and uses it to work out difference in \(y\)-coordinates of \(P\) and \(Q\) | ||
| \(\dfrac{1 + r}{2} = 5\) or (\(r =\)) \(5 \times 2 - 1\) or (\(r =\)) \(1 + 4 + 4\) or (\(r =\)) 9 | M1 | oe equation or calculation eg \(\dfrac{r - 1}{2} + 1 = 5\) or (\(r =\)) \(5 + 5 - 1\) may be seen on a sketch implied by \(Q\,(9, 6)\) implied by difference in \(x\)-coordinates of \(P\) and \(Q\) is 8 |
| (\(y_Q - y_P =\)) \(2 \times (\text{their } 9 - 1)\) or 16 | M1 | oe eg \(2 \times 8\) may be seen on a sketch as difference of \(y\)-coordinates of \(P\) and \(Q\) their 9 must be a value their 9 cannot be 1 or 5 |
| \(6 - 2 \times (\text{their } 9 - 1)\) | M1dep | oe eg \(6 - 16\) dep on 2nd M1 |
| \(-10\) | A1 | |
| Alternative method 4 Works out difference in \(y\)-coordinates of \(P\) and midpoint (\(M\)) and uses it to work out \(y\)-coordinate of \(M\) | ||
| (\(y_M - y_P =\)) \(2 \times (5 - 1)\) or 8 | M1 | oe eg \(2 \times 4\) implied by difference in \(y\)-coordinates of \(P\) and \(M\) is 8 seen on a sketch |
| (\(y\)-coordinate of \(M\) =) \(6 - 2 \times (5 - 1)\) or \(-2\) | M1dep | oe eg \(6 - 8\) may be seen on a sketch |
| their \(-2 - 2 \times (5 - 1)\) | M1dep | oe eg \(-2 - 8\) or \(6 - 2 \times (5 - 1) - 2 \times (5 - 1)\) |
| \(-10\) | A1 | |
| Alternative method 5 Works out an expression for \(y\)-coordinate of midpoint (\(M\)) and uses it to work out gradient expression for \(PM\) | ||
| (\(y\)-coordinate of \(M\) =) \(\dfrac{6 + k}{2}\) | M1 | oe may be seen on a sketch |
| \(\dfrac{\frac{6 + k}{2} - k}{5 - 1}\) | M1dep | oe expression eg \(\dfrac{k - \frac{6 + k}{2}}{1 - 5}\) or \(\dfrac{6 - k}{8}\) |
| \(\dfrac{\frac{6 + k}{2} - k}{5 - 1} = 2\) | M1dep | oe equation eg \(6 - k = 16\) |
| \(-10\) | A1 | |
Additional guidance
| Up to M3 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| In all alts a fully correct 2nd M implies the 1st M | |
| NOTE In Alts 4 and 5 the value of \(r\) is not needed |