Foundation November 2021 Paper 2 Q31
31 A straight line
has gradient 6
and
passes through the point (3, 19)
Work out the equation of the line.
Give your answer in the form \(\quad y = mx + c\) [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(6 \times 3 + c = 19\) | M1 | oe eg \(18 + c = 19\) |
| \((c =)\ 19 - 6 \times 3\) or \((c =)\) 1 | M1dep | oe implied by (0, 1) |
| \(y = 6x + 1\) | A1 | SC1 \(y = 6x + c \quad c \ne 1\) |
| Alternative method 2 | ||
| \(y - 19 = 6(x - 3)\) | M1 | oe |
| \(y - 19 = 6x - 18\) | M1dep | oe correct equation with brackets expanded |
| \(y = 6x + 1\) | A1 | SC1 \(y = 6x + c \quad c \ne 1\) |
Additional guidance
| Allow \(y = 6 \times x + 1\) | |
| \(6x + 1\) on answer line, \(y = 6x + 1\) seen in working | M1M1A1 |
| \(6x + 1\) on answer line, \(y = 6x + 1\) not seen in working | M1M1A0 |
| \(m = 6\), \(c = 1\) on answer line, \(y = 6x + 1\) seen in working | M1M1A1 |
| \(m = 6\), \(c = 1\) | M1M1A0 |
| \(y = mx + 1\) | M1M1A0 |
| Allow embedded value for \(c\) eg \(19 = 6 \times 3 + 1\) | M1M1A0 |
| \(y = 6x + c\) | SC1 |
| \(y = 6x\) | SC1 |
| \(6x + c\) on answer line with \(c \ne 1\), \(y = 6x + c\) seen in working | SC1 |
| \(6x + c\) on answer line with \(c \ne 1\), \(y = 6x + c\) not seen in working | M0M0A0 |