Higher June 2019 Paper 3 Q12
12 A straight line
has gradient 4
and
passes through the point (5, 23)
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 | ||
| \(4 \times 5 + c = 23\) | M1 | oe \(20 + c = 23\) |
| \(c = 3\) | A1 | implied by (0, 3) or 3 shown as \(y\)-axis intercept |
| \(y = 4x + 3\) | A1 | SC1 \(\ y = 4x + c \quad c \ne 3\) |
| Alternative method 2 | ||
| \(y - 23 = 4(x - 5)\) | M1 | oe |
| \(y - 23 = 4x - 20\) | M1dep | |
| \(y = 4x + 3\) | A1 | SC1 \(\ y = 4x + c \quad c \ne 3\) |
Additional guidance
| If 3 is clearly linked to \(c\) in \(y = mx + c\) condone M1A1 | |
| \(4x + 3\) on answer line, \(y = 4x + 3\) seen in working | M1A1A1 |
| \(4x + 3\) on answer line, \(y = 4x + 3\) not seen in working | M1A1A0 |
| \(m = 4,\ c = 3\) on answer line, \(y = 4x + 3\) seen in working | M1A1A1 |
| \(m = 4,\ c = 3\) | M1A1A0 |
| \(y = mx + 3\) | M1A1A0 |
| \(23 = 4 \times 5 + 3\) embedded value for \(c\) | M1A0A0 |
| \(4x + c\) on answer line with \(c \ne 3\) | M0A0A0 |