Higher January 2020 Paper 2 Q19
19
(a) Write down an equation of a line that is parallel to the line with equation \(\;y = 7 - 4x\) (1)
The line L passes through the points with coordinates \((-3, 1)\) and \((2, -2)\)
(b) Find an equation of the line that is perpendicular to L and passes through the point with coordinates \((-6, 4)\)
Give your answer in the form \(\;ax + by + c = 0\;\) where \(a\), \(b\) and \(c\) are integers. (4)
Give your answer in the form \(\;ax + by + c = 0\;\) where \(a\), \(b\) and \(c\) are integers. (4)
| Scheme | Marks |
|---|---|
| \(y = -4x + k\) (oe) | B1 |
| (1) |
Notes
B1: for \(y = -4x\) or \(y = -4x + k\) where \(k\) is any numerical value \(k \neq 7\)
Could be written in another form e.g. \(3y + 12x = 20\)
| Scheme | Marks |
|---|---|
| \(m = \dfrac{-2 - 1}{2 - -3}\) or \(m = \dfrac{1 - -2}{-3 - 2}\) or \(-\dfrac{3}{5}\) or −0.6 | M1 |
| \(m_p = \dfrac{5}{3}\) | M1ft |
\(4 = \dfrac{5}{3}(-6) + c\) oe eg \(4 = -10 + c\) (\(c = 14\)) \(y - 4 = \dfrac{5}{3}(x - -6)\) | M1ft |
| Eg \(5x - 3y + 42 = 0\) | A1 |
| (4) | |
| (5 marks) |
Notes
M1: for using \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\)
M1ft: for using \(m_1 \times m_2 = -1\)
M1ft: dep on previous M1 for substituting \((-6, 4)\) into linear equation formula
\(4 = \dfrac{5}{3}x + c\) to find value of \(c\) or
\(y = \dfrac{5}{3}x + 14\) or \(y = 1.66...x + 14\)
A1: for correct simplified equation where all values are integers
\(10x - 6y + 84 = 0\) or
\(3y = 5x + 42\) oe