Standard +0.3 This is a straightforward multi-step problem requiring finding the line equation, substituting into the quadratic, solving for intersection points, and applying the distance formula. All techniques are standard A-level procedures with no novel insight required, making it slightly easier than average.
7 Determine the exact distance between the two points at which the line through ( 4,5 ) and ( \(6 , - 1\) ) meets the curve \(y = 2 x ^ { 2 } - 7 x + 1\).
Three term quadratic seen or implied by correct \(x\)-values
\(x = -2, 4\)
A1
cao
When \(x = 4\), \(y = 5\); when \(x = -2\), \(y = 23\)
A1
Both \(y\)-coordinates seen
Distance between \((4,5)\) and \((-2, 23)\): \(\sqrt{(-2-4)^2 + (23-5)^2}\)
M1
Uses distance formula for their points (not given points)
\(= 6\sqrt{10}\)
A1
Must be exact (allow \(\sqrt{360}\) oe)
## Question 7:
| Answer | Mark | Guidance |
|--------|------|----------|
| Line through $(4,5)$ and $(6,-1)$ has gradient $\frac{-1-5}{6-4} = -3$, so equation is $y = 17 - 3x$ | M1 | Attempt to find equation of line using correct gradient formula |
| Points of intersection: $2x^2 - 7x + 1 = 17 - 3x$ | M1 | Eliminating one variable |
| $2x^2 - 4x - 16 = 0$ | M1 | Three term quadratic seen or implied by correct $x$-values |
| $x = -2, 4$ | A1 | cao |
| When $x = 4$, $y = 5$; when $x = -2$, $y = 23$ | A1 | Both $y$-coordinates seen |
| Distance between $(4,5)$ and $(-2, 23)$: $\sqrt{(-2-4)^2 + (23-5)^2}$ | M1 | Uses distance formula for their points (not given points) |
| $= 6\sqrt{10}$ | A1 | Must be exact (allow $\sqrt{360}$ oe) |
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7 Determine the exact distance between the two points at which the line through ( 4,5 ) and ( $6 , - 1$ ) meets the curve $y = 2 x ^ { 2 } - 7 x + 1$.
\hfill \mbox{\textit{OCR MEI Paper 1 2023 Q7 [7]}}