Higher June 2023 Paper 2 Q24
24 \(ABCD\) is a kite with \(AB = AD\) and \(CB = CD\)
\(A\) is the point with coordinates \((-2, 10)\)
\(B\) is the point with coordinates \(\left(-\dfrac{27}{5}, 4\right)\)
\(C\) is the point with coordinates \((4, -5)\)
Work out the coordinates of \(D\)
(6)
| Scheme | Marks |
|---|---|
| \(\dfrac{-5 - 10}{4 - -2}\;\left(= -\dfrac{5}{2}\right)\) | M1 |
\(y - 10 = -\dfrac{5}{2}(x + 2)\) oe eg \(y = -\dfrac{5}{2}x + 5\) or \(y - -5 = -\dfrac{5}{2}(x - 4)\) oe or \(5x + 2y = 10\) oe | M1 |
\(y - 4 = \dfrac{2}{5}\left(x - -\dfrac{27}{5}\right)\) oe or \(4 = \dfrac{2}{5}\left(-\dfrac{27}{5}\right) + c\;\left(y = \dfrac{2}{5}x + 6.16\right)\) \(\dfrac{4 - y}{-\frac{27}{5} - x} = \dfrac{2}{5}\) oe or \(5y - 2x = \dfrac{154}{5}\) oe | M1 |
solves \(-\dfrac{5}{2}x + 5 = \dfrac{2}{5}x + 6.16\) oe eg \(10x + 4y = 20\) and \(-10x + 25y = 154\) oe, with operation of addition or \(25x + 10y = 50\) and \(-4x + 10y = 61.6\) oe, with operation of subtraction or \(x = \dfrac{5}{2}y - \dfrac{154}{10}\) oe or \(y = \dfrac{2}{5}x + \dfrac{154}{25}\) oe substituted in other equation | M1 |
| Coordinates of intersection of \(AC\) and \(BD\): \(x = -\dfrac{2}{5}\), \(y = 6\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: (4.6, 8) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: A correct method to find the gradient of \(AC\)
M1: ft (if M1 scored) correct equation of \(AC\)
M1: ft (if first M1 scored)
equation of \(BD\) or correct equation using gradient of \(BD\)
M1: Solve equation OR
Solve simultaneously the correct equations of lines of \(AC\) and \(BD\) or correct equation from gradient or other correct equation.
If elimination: same coefficient of \(x\) or \(y\) with suitable sign used to eliminate.
If substitution: \(x\) or \(y\) substituted into other equation.
M1: oe value of \(x\) and \(y\) at intersection of \(AC\) and \(BD\)
A1: oe coordinates of \(D\)
24 ALT (working with \(AD = AB\), \(CD = CB\) or gradients)
| Scheme | Marks |
|---|---|
eg \((10 - 4)^2 + \left(-2 + \dfrac{27}{5}\right)^2\) (= 47.56) (\(AB\) = 6.896…) or eg \((-5 - 4)^2 + \left(4 + \dfrac{27}{5}\right)^2\) (= 169.36) (\(CB\) = 13.013…) or eg \(\dfrac{-5 - 10}{4 - -2}\) or \(\dfrac{4 - y}{-\frac{27}{5} - x}\) oe | M1 |
eg \((y - 10)^2 + (x + 2)^2 = (10 - 4)^2 + \left(-2 + \dfrac{27}{5}\right)^2\) or eg \((y + 5)^2 + (x - 4)^2 = (-5 - 4)^2 + \left(4 + \dfrac{27}{5}\right)^2\) or \(\dfrac{-5 - 10}{4 - -2} \times \dfrac{4 - y}{-\frac{27}{5} - x} = -1\) oe eg \(-60 + 15y = 6x + 32.4\) | M1 |
| eg \(2x - 5y = -30.8\) or \(x = 2.5y - 15.4\) or \(y = 0.4x + 6.16\) oe | M1 |
eg \((y - 10)^2 + (2.5y - 15.4 + 2)^2 = (10 - 4)^2 + \left(-2 + \dfrac{27}{5}\right)^2\) eg \((0.4x + 6.16 + 5)^2 + (x - 4)^2 = (-5 - 4)^2 + \left(4 + \dfrac{27}{5}\right)^2\) | M1 |
| \(7.25y^2 - 87y + 232 = 0\) oe or \(1.16x^2 + 0.928x - 28.8144 = 0\) oe | M1 |
| (4.6, 8) | A1 |
| (6 marks) |
Notes
M1: A correct method to find \(AB^2\) or \(CB^2\)
or
\(AB\) or \(CB\)
or
a correct gradient expression for \(AC\) or \(DB\)
M1: Using \(D\) (\(x\), \(y\)) form a correct equation
\(AD^2 = AB^2\)
or
\(CD^2 = CB^2\)
or
gradients \(AC \times DB = -1\) (Using \(D\) (\(x\), \(y\)))
M1: uses rearrangement or solving simultaneous equations to find a correct 3 term linear equation
M1: uses substitution to obtain a correct quadratic equation in one unknown
M1: for a 3 term quadratic that can be used to find the value of \(x\) or the value of \(y\) at \(D\)
A1: oe coordinates of \(D\)