June 2016 Paper 1 Q11
11 \(A\left(1\frac{1}{5}, 3\frac{4}{5}\right)\), \(B\left(2, 1\frac{4}{5}\right)\) and \(C(5, 3)\) are points on a coordinate grid.
Show that the line segments \(AB\) and \(BC\) are perpendicular. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Correct method for finding the gradient of either \(AB\) or \(BC\) | M1 | \(\dfrac{1\frac{4}{5} - 3\frac{4}{5}}{2 - 1\frac{1}{5}}\) or \(\dfrac{1.8 - 3.8}{2 - 1.2}\) or \(\dfrac{3 - 1\frac{4}{5}}{5 - 2}\) or \(\dfrac{3 - 1.8}{5 - 2}\) |
| Correct method for finding the gradient of \(AB\) and \(BC\), and at least one of the gradients correct | M1 | \(\dfrac{-10}{4}\) or \(\dfrac{-5}{2}\) or \(-2\frac{1}{2}\) or \(-2.5\) or \(\dfrac{-2}{\frac{4}{5}}\) or \(\dfrac{-2}{0.8}\) or \(\dfrac{6}{15}\) or \(\dfrac{2}{5}\) or \(\dfrac{4}{10}\) or 0.4 or \(\dfrac{1\frac{1}{5}}{3}\) or \(\dfrac{1.2}{3}\) |
| product \(= -1\) clearly shown or clearly showing that one is the negative reciprocal of the other | A1 | |
| Alternative method 2 | ||
| \(AB^2 = (2 - 1.2)^2 + (1.8 - 3.8)^2\) or \(BC^2 = (5 - 2)^2 + (3 - 1.8)^2\) or \(AC^2 = (5 - 1.2)^2 + (3 - 3.8)^2\) | M1 | oe |
| \(AB^2 = (0.8)^2 + (-2)^2\) and \(BC^2 = (3)^2 + (1.2)^2\) and \(AC^2 = (3.8)^2 + (-0.8)^2\) | M1 | oe all expressions simplified but not necessarily evaluated |
| \(AB^2 + BC^2 = 4.64 + 10.44 = 15.08\) and \(AC^2 = 15.08\) | A1 | showing equal values is sufficient |
Additional guidance
The \(\dfrac{y\text{-step}}{x\text{-step}}\) calculation, for the first M1, might be on the diagram
The expressions for the gradients can be either way round eg \(\dfrac{3.8 - 1.8}{1.2 - 2} = \dfrac{2}{-0.8}\)
Both gradients numerically correct but both with the wrong sign eg \(\dfrac{10}{4}\) and \(\dfrac{-6}{15}\) scores SC1
For the 2nd M mark we are looking for \(m/n\) where \(m\) and \(n\) are integers, mixed numbers, fractions, improper fractions or decimals eg \(\dfrac{6/5}{3}\)
The gradients for each line can be found by forming two simultaneous equations. It will be very rare, but check their working if you see this method.