22 \(\mathbf{C}\) is a circle with centre \((0, 0)\) The straight line with equation \(3x - 2y = 52\) is the tangent to \(\mathbf{C}\) at the point \(P\).
Find the coordinates of \(P\). (4)
Mark scheme
Answer
Mark
Mark scheme
(12, −8)
P1
for process to rearrange the equation to give \(y\) in terms of \(x\), eg \(y = \dfrac{3x - 52}{2}\) or \(y = \dfrac{3}{2}x - 26\) or \(m = \dfrac{3}{2}\)
P1
for process to find gradient of \(OP\), eg \(-1 \div \text{``}\dfrac{3}{2}\text{''}\ \left(= -\dfrac{2}{3}\right)\) or \(-1 \div [m]\)
P1
(dep on equation of the form \(y = \dfrac{-1}{[m]}x\) for the radius may be implied in subsequent working) for starting to solve \(3x - 2y = 52\) with \(y = \dfrac{-1}{[m]}x\) simultaneously to find the value of \(x\) or \(y\) eg substituting \(y = \dfrac{-2}{3}x\) or \(y = \dfrac{-1}{[m]}x\) into \(y = \dfrac{3}{2}x - 26\) or \(3x - 2y = 52\) or an equation of the form \(y = \dfrac{3}{2}x + c\) eg \(-\dfrac{2}{3}x = \dfrac{3}{2}x - 26\) or \(3x - 2\left(-\dfrac{2}{3}x\right) = 52\) or \(-\dfrac{2}{3}x = \dfrac{3}{2}x + c\) or \(-\dfrac{1}{[m]}x = \dfrac{3}{2}x - 26\) or \(3x - 2\left(-\dfrac{1}{[m]}x\right) = 52\) or \(-\dfrac{1}{[m]}x = \dfrac{3}{2}x + c\)
A1
cao
Additional guidance
Condone an incorrect value for the \(y\) intercept (\(= c\))
Where \([m]\) is clearly identified as the gradient of the straight line \(3x - 2y = 52\) Allow 0.66(6…) or 0.67 for \(\dfrac{2}{3}\) throughout
Where \([m]\) is clearly identified as the gradient of the straight line \(3x - 2y = 52\)
Can be done by elimination Award P1 for a correct method to eliminate \(x\) or \(y\): coefficient of \(x\) or \(y\) the same and correct operator to eliminate selected variable eg
for substitution of a rearranged equation into a correct equation to form an equation in one variable. eg \(3x^2 + 2(2 - 3x)^2 = 44\) or \(3\left(\dfrac{2 - y}{3}\right)^2 + 2y^2 = 44\)
M1
(dep on first M1) for multiplying out all brackets and collecting terms to form a simplified three term quadratic in any form of \(ax^2 + bx + c\ (= 0)\) where at least 2 coefficients (\(a\), \(b\), \(c\)) are correct
(dep on first M1) for a suitable method to solve their 3 term quadratic using any correct method,
for factorising, eg \((7x + 6)(x - 2)\) or \((7x + 6)(3x - 6)\) or \((21x + 18)(x - 2)\) or \((7y - 32)(y + 4)\)
or correct use of formula, eg \(\dfrac{8 \pm \sqrt{(-8)^2 - 4 \times 7 \times -12}}{2 \times 7}\) or \(\dfrac{4 \pm \sqrt{(-4)^2 - 4 \times 7 \times -128}}{2 \times 7}\) or completing the square
M1
(dep on first M1) for substituting their 2 found values of \(x\) or \(y\) in a suitable equation or (dep on first M1) for one correct pair of values following from a correct quadratic
A1
for \(x = -\dfrac{6}{7}\) oe, \(y = \dfrac{32}{7}\) oe and \(x = 2\), \(y = -4\)
Additional guidance
Allow \((\pm 2 \pm 3x)\) for \((2 - 3x)\) (or \(\left(\dfrac{\pm 2 \pm y}{3}\right)\) for \(\left(\dfrac{2 - y}{3}\right)\)) Implied by a correct equation (simplified or unsimplified) in terms of \(x\) or \(y\) eg \(3x^2 + 2(4 - 12x + 9x^2) = 44\) or \(3x^2 + 8 - 24x + 18x^2 = 44\) or \(21x^2 - 24x = 36\)
Look out for signs reversed The quadratic does not have to equal 0, ie accept \(21x^2 - 24x = 36\)
Can be implied by both \(x\) values or both \(y\) values correct (condone incorrect labelling) if the quadratic is correct If using the quadratic formula (condone one sign error, omission of brackets around the \(b\) in the \(b^2 - 4ac\) and the fraction line not being under the \(-\) in the \(-b\). Allow some simplification – as far as eg \(\dfrac{8 \pm \sqrt{64 + 336}}{14}\) or if factorising allow brackets which expand to give 2 out of 3 terms correct for their quadratic
Condone substitution into their \((\pm 2 \pm 3x)\) or \(\left(\dfrac{\pm 2 \pm y}{3}\right)\)
Allow \(-0.85(7\ldots)\) or \(-0.86\) for \(-\dfrac{6}{7}\) Allow 4.57(1…) for \(\dfrac{32}{7}\) If values of \(x\) or \(y\) are incorrect then working must be shown
Accept as coordinates Assume correct pairing unless clearly incorrect eg \(\left(-\tfrac{6}{7}, -4\right), \left(2, \tfrac{32}{7}\right)\) Allow \(-0.85(7\ldots)\) or \(-0.86\) for \(-\dfrac{6}{7}\) Allow 4.57(1…) for \(\dfrac{32}{7}\) If an answer is shown in the range in working and then incorrectly rounded award full marks
A correct answer with no supportive working gets 0 marks
for correct substitution for \(y^2\) or \(x^2\), eg \((7 - 2x)^2 = 3x^2 + 4\) OR for correct rearrangement and expansion of \((7 - 2x)^2\) to obtain 4 terms with all correct without considering signs or for 3 terms out of 4 correct with correct signs and substitution eg \((7 - 2x)^2 = 49 - 14x - 14x + 4x^2\) and \(49 - 14x - 14x + 4x^2 = 3x^2 + 4\)
M1
for method to write a correct simplified equation eg \(x^2 - 28x + 45\ (= 0)\)
M1
for a method to solve a correct quadratic eg \(\dfrac{28 \pm \sqrt{(-28)^2 - 4 \times 1 \times 45}}{2 \times 1}\) or \(\dfrac{28 \pm \sqrt{604}}{2}\) or \(14 \pm \sqrt{151}\) or \((x - 14)^2 - 14^2 + 45 = 0\) oe
A1
\(x = 26.2\) to 26.3, \(y = -45.6\) to \(-45.5\) and \(x = 1.7\) to 1.712, \(y = 3.5\) to 3.6
Additional guidance
NB \(49 - 28x\) or \(-28x + 4x^2\) can be considered 3 terms out of 4 correct with correct signs
The quadratic does not have to equal 0, ie accept \(x^2 - 28x = -45\)
Can be implied by both \(x\) values correct or both \(y\) values correct
Answers must be correctly paired (May be in the body of the working) If answers are given in the range in working and then rounded incorrectly award full marks
16 1 cup of tea and 1 cup of coffee cost £4.50 3 cups of tea and 1 cup of coffee cost £8.50
Work out the total cost of 4 cups of tea and 3 cups of coffee. You must show all your working. (5)
Mark scheme
Answer
Mark
Mark scheme
15.5(0)
P1
for a correct process to find the cost of some teas only or some coffees only eg (2 teas =) \(8.50 - 4.50\ (= 4)\) or (2 coffees =) \(3 \times 4.50 - 8.50\ (= 5)\)
P1
for a process to find the cost of one tea or four teas, eg (1 tea =) \(\text{``}4\text{''} \div 2\ (= 2)\) or (4 teas =) \(\text{``}4\text{''} \times 2\ (= 8)\)
P1
for a process to find cost of one coffee or three coffees, eg (1 coffee =) \(4.50 - \text{``}2\text{''}\ (= 2.50)\) or \(8.50 - \text{``}2\text{''} \times 3\ (= 2.50)\) or \(1.5 \times \text{``}5\text{''}\ (= 7.50)\)
P1
for a complete process, eg \(\text{``}2\text{''} \times 4 + \text{``}2.50\text{''} \times 3\)
A1
cao
Additional guidance
Use of simultaneous equations – marks awarded at the stages shown in the scheme.
