Easy -1.2 This is a straightforward application of y - y₁ = m(x - x₁) followed by simple rearrangement and finding intercepts by substituting x=0 and y=0. It requires only basic recall of the point-gradient formula with minimal algebraic manipulation, making it easier than average for A-level.
Find the equation of the line with gradient \(-2\) which passes through the point \((3, 1)\). Give your answer in the form \(y = ax + b\).
Find also the points of intersection of this line with the axes. [3]
M1 for \(y - 1 = -2(x - 3)\) or \(l = -2 \times 3 + c\) oe
\((0, 7)\) and \((3.5, 0)\) oe or ft their \(y = -2x + c\)
1
condone lack of brackets and eg \(y = 7, x = 3.5\) or ft isw but 0 for poor notation such as \((3.5, 7)\) and no better answers seen
[3]
$y = -2x + 7$ isw | 2 | M1 for $y - 1 = -2(x - 3)$ or $l = -2 \times 3 + c$ oe
$(0, 7)$ and $(3.5, 0)$ oe or ft their $y = -2x + c$ | 1 | condone lack of brackets and eg $y = 7, x = 3.5$ or ft isw but 0 for poor notation such as $(3.5, 7)$ and no better answers seen
| [3] |
Find the equation of the line with gradient $-2$ which passes through the point $(3, 1)$. Give your answer in the form $y = ax + b$.
Find also the points of intersection of this line with the axes. [3]
\hfill \mbox{\textit{OCR MEI C1 2012 Q1 [3]}}