June 2017 Paper 1 Q4
4 A straight line passes through the points (\(-4\), 7), (6, \(-5\)) and (8, \(t\))
Use an algebraic method to work out the value of \(t\).
You must show your working. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| Intention to work out gradient or reciprocal of gradient or Intention to work out the equation of the straight line | M1 | Condone one sign error in the calculation eg \(\dfrac{-5 - 7}{6 - -4}\) or \(\dfrac{-5 - t}{6 - 8}\) or \(\dfrac{7 - t}{-4 - 8}\) or \(-1.2\) oe eg \(7 = -4m + c\) or \(-5 = 6m + c\) eg \(y - 7 = m(x - -4)\) |
| A correct value for \(m\) or a correct expression for \(m\) and an expression to calculate the value of \(t\) or the value of \(c\) or \(m = -1.2\) and \(c = 2.2\) | M1dep | eg. (\(m\) =) \(\dfrac{7 - -5}{-4 - 6}\) or (\(m\) =) \(\dfrac{-6}{5}\) oe and eg \(\dfrac{t - -5}{8 - 6} = \dfrac{-5 - 7}{6 - -4}\) or \(t = \dfrac{-6}{5}(8) + \left(7 - \dfrac{24}{5}\right)\) or \(7 = \dfrac{-6}{5}(-4) + c\) or \(-5 = \dfrac{-6}{5}(6) + c\) |
| (\(t\) =) \(-7.4\) or \(-7\frac{2}{5}\) or \(\dfrac{-37}{5}\) | A1 | |
| Alternative method 2 | ||
| \(-4\) to 6 is \(+10\) and 7 to \(-5\) is \(-12\) | M1 | oe Condone a sign error |
| 6 to 8 is \(+2\) and \(-5\) to \(t\) is \(\dfrac{-12}{5}\) | M1 | oe |
| (\(t\) =) \(-7.4\) or \(-7\frac{2}{5}\) or \(\dfrac{-37}{5}\) | A1 | |
| Alternative method 3 | ||
| \(\sqrt{[(-4 - 6)^2 + (7 - -5)^2]}\) (\(= \sqrt{244}\)) and stating \(-4\) to 6 is 10 and 6 to 8 is 2 | M1 | Correct use of Pythagoras and identifying the correct displacements |
| \(\sqrt{[(6 - 8)^2 + (-5 - t)^2]} = (\sqrt{244}) \div 5\) | M1 | ft their 244 |
| (\(t\) =) \(-7.4\) or \(-7\frac{2}{5}\) or \(\dfrac{-37}{5}\) | A1 | |
Additional guidance
\(-7.4\) seen on answer line is 3 marks
\(-7.4\) seen in the working but sign error on answer line is 3 marks
'Algebraic method' means the question must not be done graphically ... although a diagram is fine when used to do the gradient calculations
\(\dfrac{t - -5}{8 - 6} = \dfrac{t - 7}{8 - -4}\) seen implies M1 M1
Look at any diagram they may have drawn for evidence of the alt 2 method
\(\dfrac{7 - -5}{-4 - 6}\) (correct expression) \(= 1.2\) (error) followed by \(7 = (1.2)(-4) + c\) scores M1 M1 but will not lead to a correct final answer, so A0
\(m = -1.2\), but they use 1.2 instead ... \(7 = 1.2(-4) + c\) giving \(c = 11.8\) is M1 M1 A0
\(m = -1.2\), then \(t = -1.2 + 11.8 = 2.2\) scores M1 M1 A0 because this is a correct method for calculating \(c\), and so scores the 2nd M1, even though they think they are calculating \(t\)
\(m = \dfrac{-5 - 7}{6 - -4} = \dfrac{-12}{10}\quad \dfrac{-12}{10} \times 2 = \dfrac{-24}{20} = \dfrac{-6}{5} = -1.2\) so \(t = -5 - 1.2 = -6.2\) M1 M1 A0
... because the only error is \(\dfrac{-12}{10} \times 2 = \dfrac{-24}{20}\) ... if this had been \(-2.4\) then \(t = -7.4\)