Easy -1.3 This is a straightforward application of the gradient formula or point-slope form requiring only substitution into m = (y₂-y₁)/(x₂-x₁) or y = mx + c. It's a single-step calculation with no problem-solving element, making it easier than average but not trivial since students must recall and apply the correct formula.
condone omission of brackets; or M1 for correct method for eqn of line and \(x = 5\) subst in their eqn and evaluated to find \(k\); or M1 for both of \(y - k = 3(x - 5)\) oe and \(y - (-5) = 3(x - 1)\) oe
M1 for \(\frac{k-(-5)}{5-1} = 3\) or other correct use of gradient eg triangle with 4 across, 12 up
Question 3(i)
Answer
Marks
Guidance
\(4/3\) isw
2
condone \(\pm4/3\); M1 for numerator or denominator correct or for \(\frac{3}{4}\) or \(\frac{1}{3/4}\) oe or for \(\left(\frac{16}{9}\right)^{\frac{1}{2}}\) soi
Question 3(ii)
Answer
Marks
Guidance
\(\frac{2a}{c^5}\) or \(2ac^{-5}\)
3
B1 for each 'term' correct; mark final answer; if B0, then SC1 for \((2ac^2)^3 = 8a^3c^6\) or \(72a^3c^6\) seen
Question 4(i)
Answer
Marks
Guidance
\((10, 4)\)
2
0 for \((5, 4)\); otherwise 1 for each coordinate
Question 4(ii)
Answer
Marks
Guidance
\((5, 11)\)
2
0 for \((5, 4)\); otherwise 1 for each coordinate
$y = 7$ or $(5, 7)$ | 2 | condone omission of brackets; or M1 for correct method for eqn of line and $x = 5$ subst in their eqn and evaluated to find $k$; or M1 for both of $y - k = 3(x - 5)$ oe and $y - (-5) = 3(x - 1)$ oe
M1 for $\frac{k-(-5)}{5-1} = 3$ or other correct use of gradient eg triangle with 4 across, 12 up
## Question 3(i)
$4/3$ isw | 2 | condone $\pm4/3$; M1 for numerator or denominator correct or for $\frac{3}{4}$ or $\frac{1}{3/4}$ oe or for $\left(\frac{16}{9}\right)^{\frac{1}{2}}$ soi | M1 for just $-4/3$; allow M1 for $\sqrt{16} = 4$ and $\sqrt{9} = 3$ soi; condone missing brackets
## Question 3(ii)
$\frac{2a}{c^5}$ or $2ac^{-5}$ | 3 | B1 for each 'term' correct; mark final answer; if B0, then SC1 for $(2ac^2)^3 = 8a^3c^6$ or $72a^3c^6$ seen | condone $a^{1}$; condone multiplication signs but **0** for addition signs
## Question 4(i)
$(10, 4)$ | 2 | **0** for $(5, 4)$; otherwise 1 for each coordinate | ignore accompanying working / description of transformation; condone omission of brackets
## Question 4(ii)
$(5, 11)$ | 2 | **0** for $(5, 4)$; otherwise 1 for each coordinate | ignore accompanying working / description of transformation; condone omission of brackets
A line has gradient 3 and passes through the point $(1, -5)$. The point $(5, k)$ is on this line. Find the value of $k$. [2]
\hfill \mbox{\textit{OCR MEI C1 2011 Q2 [2]}}