Higher June 2025 Paper 3 Q22
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)
| 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
\[\begin{array}{r} 9x - 6y = 156 \\ \underline{+\ 4x + 6y = 0\phantom{00}} \\ (13x \phantom{{}+6y} = 156) \end{array} \qquad \begin{array}{r} 6x - 4y = 104 \\ \underline{-\ 6x + 9y = 0\phantom{00}} \\ (-13y = 104) \end{array}\]