Higher January 2020 Paper 1 Q22
22 Triangle \(HJK\) is isosceles with \(\;HJ = HK\;\) and \(\;JK = \sqrt{80}\)
\(H\) is the point with coordinates \((-4, 1)\)
\(J\) is the point with coordinates \((j, 15)\) where \(\;j \lt 0\)
\(K\) is the point with coordinates \((6, k)\)
\(M\) is the midpoint of \(JK\).
The gradient of \(HM\) is 2
Find the value of \(j\) and the value of \(k\).
(6)
| Scheme | Marks |
|---|---|
| gradient of \(JK = -0.5\) or \(m \times 2 = -1\) | M1 |
| \(\dfrac{k - 15}{6 - j} = -\dfrac{1}{2}\) or \(2k - j = 24\) or \(j = 2k - 24\) or \(k = \dfrac{j + 24}{2}\) oe | M1 |
\((j - 6)^2 + (k - 15)^2 = 80\) oe or \(\left(\dfrac{j + 6}{2}, \dfrac{k + 15}{2}\right)\) oe or \((j + 4)^2 + 196 = 100 + (k - 1)^2\) oe | M1 |
eg \(3k^2 - 78k + 495 = 0\) oe or \(5j^2 - 60j - 140 = 0\) oe or \(5k^2 - 150k + 1045 = 0\) oe or \(3j^2 - 12j - 36 = 0\) oe or gradient \(HM\): eg \(\dfrac{\frac{k + 15}{2} - 1}{\frac{j + 6}{2} + 4} = 2\) or \(k = 2j + 15\) or \(j = \dfrac{k - 15}{2}\) oe | M1 |
eg \((k - 15)(k - 11)\;(= 0)\) or \(\dfrac{78 \pm \sqrt{(-78)^2 - 4 \times 3 \times 495}}{2 \times 3}\) or \((k - 13)^2 - 169 + 165\;(= 0)\) or eg \((j - 6)(j + 2)\;(= 0)\) or \(\dfrac{12 \pm \sqrt{(-12)^2 - 4 \times 3 \times -36}}{2 \times 3}\) or \((j - 2)^2 - 4 - 12\;(= 0)\) | M1 |
| \(j = -2\), \(k = 11\) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for finding the gradient of \(JK\) using \(m_1 \times m_2 = -1\)
M1: for expressing the gradient of \(JK\) in terms of \(j\) and \(k\) or a correct equivalent equation
M1: for finding equation of \(JK\) in terms of \(j\) and \(k\)
or
for finding the midpoint of \(M\)
or
for equating length \(HJ\) with length \(HK\)
M1: (dep on M3) writing a correct quadratic expression in the form \(ax^2 + bx + c\;(= 0)\) (allow \(ax^2 + bx = c\))
or
A correct equation for the gradient of \(HM\) in terms of \(j\) and \(k\) or a correct equivalent equation
ALT
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{j + 6}{2}, \dfrac{k + 15}{2}\right)\) oe | M1 |
\(\dfrac{\frac{k + 15}{2} - 1}{\frac{j + 6}{2} + 4} = 2\) or \(k - 2j = 15\) or \(k = 2j + 15\) or \(j = \dfrac{k - 15}{2}\) oe | M1 |
| \((j - 6)^2 + (k - 15)^2 = 80\) oe or \((j + 4)^2 + 196 = 100 + (k - 1)^2\) oe | M1 |
| E.g. \(5j^2 - 12j - 44 = 0\) or \(3j^2 + 48j + 84 = 0\) oe or E.g. \(5k^2 - 174k + 1309 = 0\) or \(3k^2 + 6k - 429 = 0\) oe | M1 |
E.g. \((5j - 22)(j + 2)\;(= 0)\) or \(\dfrac{12 \pm \sqrt{(-12)^2 - 4 \times 5 \times -44}}{2 \times 5}\) or \((j + 8)^2 - 64 + 28\;(= 0)\) or E.g. \((5k - 119)(k - 11)\;(= 0)\) or \(\dfrac{174 \pm \sqrt{(-174)^2 - 4 \times 5 \times 1309}}{2 \times 5}\) or \((k + 1)^2 - 1 - 143\;(= 0)\) | M1 |
| \(j = -2\), \(k = 11\) | A1 |
Notes
M1: for finding the midpoint of \(M\)
M1: for expressing the gradient of \(HM\) in terms of \(j\) and \(k\) or a correct equivalent equation (corrected from the printed mark scheme: it prints “gradient of \(JK\)”; the working is the gradient of \(HM\))
M1: for finding the length of \(JK\) in terms of \(j\) and \(k\)
or for equating length \(HJ\) with length \(HK\)
M1: (dep on M3) writing the correct quadratic expression in form \(ax^2 + bx + c\;(= 0)\)
allow \(ax^2 + bx = c\)
M1: (dep on M3) for a complete method to solve their 3-term quadratic equation (allow one sign error in the use of the quadratic formula)