Moderate -0.3 This is a multi-step C1 coordinate geometry question requiring finding a midpoint, identifying the gradient of a perpendicular line, and forming an equation. While it combines several standard techniques (midpoint formula, perpendicular gradients multiply to -1, point-slope form), each step is routine and the question follows a predictable structure with no novel insight required. Slightly easier than average due to straightforward arithmetic and clear signposting.
\(A\) is the point \((-2, 6)\) and \(B\) is the point \((3, -8)\). The line \(l\) is perpendicular to the line \(x - 3y + 15 = 0\) and passes through the mid-point of \(AB\). Find the equation of \(l\), giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. [7]
Answer: Midpoint of AB is \(\left(\frac{-2+3}{2}, \frac{6 + -8}{2}\right) = \left(\frac{1}{2}, -1\right)\); Gradient of given line \(= \frac{1}{3}\); Gradient of \(l = -3\); \(y + 1 = -3\left(x - \frac{1}{2}\right)\); \(6x + 2y - 1 = 0\)
Answer
Marks
Guidance
Marks: M1
A1
B1
Guidance: Correct method to find midpoint – can be implied by one correct value
Must be stated or used – just rearranging the equation is not sufficient
Additional notes: NB – "correct" answer can be found with wrong mid-pt. Check working thoroughly. Must include "= 0"
**Answer:** Midpoint of AB is $\left(\frac{-2+3}{2}, \frac{6 + -8}{2}\right) = \left(\frac{1}{2}, -1\right)$; Gradient of given line $= \frac{1}{3}$; Gradient of $l = -3$; $y + 1 = -3\left(x - \frac{1}{2}\right)$; $6x + 2y - 1 = 0$
**Marks:** M1 | A1 | B1 | B1FT | M1 | A1 | A1 [7]
**Guidance:** Correct method to find midpoint – can be implied by one correct value | | Must be stated or used – just rearranging the equation is not sufficient | Use of $m_1m_2 = -1$ (may be implied), allow for any initial non-zero numerical gradient | Correct equation for line, any non-zero numerical gradient, through their $\left(\frac{1}{2}, -1\right)$ | Correct equation in any three-term form | $k(6x + 2y - 1) = 0$ for integer k www
**Additional notes:** NB – "correct" answer can be found with wrong mid-pt. Check working thoroughly. Must include "= 0"
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$A$ is the point $(-2, 6)$ and $B$ is the point $(3, -8)$. The line $l$ is perpendicular to the line $x - 3y + 15 = 0$ and passes through the mid-point of $AB$. Find the equation of $l$, giving your answer in the form $ax + by + c = 0$, where $a$, $b$ and $c$ are integers. [7]
\hfill \mbox{\textit{OCR C1 2013 Q8 [7]}}