| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Moderate -0.8 This is a straightforward two-part question requiring basic manipulation: rearranging to find gradient from standard form, then using the perpendicular gradient rule and point-slope form. These are routine P1 techniques with no problem-solving or insight required, making it easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Attempts to make \(y\) the subject of \(3x + 5y - 7 = 0\) | M1 | Expect \(\pm 5y = \ldots\) then \(y = \ldots\); \(\frac{3}{5}\) alone is M0 |
| States \(-\frac{3}{5}\) or exact equivalent | A1 | Correct answer implies both marks; value of \(c\) need not be correct; do not allow \(-\frac{3}{5}x\) alone |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Uses perpendicular gradients rule \(\Rightarrow\) gradient \(l_2 = \frac{5}{3}\) | M1 | |
| Forms equation of \(l_2\) using \((6, -2)\): \(y + 2 = \frac{5}{3}(x - 6)\) | M1 | |
| \(y = \frac{5}{3}x - 12\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Uses perpendicular gradient rule following through on gradient from (a) | M1 | If gradient not given, follow through on their \(m\) |
| Equation of straight line with changed gradient using \((6, -2)\) | M1 | So if (a) was \(-\frac{3}{5}\), then \((y+2) = \frac{3}{5}(x-6)\) scores this. At least one bracket correct. If \(y=mx+c\) used, must find \(c\) using gradient and \((6,-2)\) |
| \(y = \frac{5}{3}x - 12\) | A1 | Allow exact equivalents e.g. \(y = \frac{10}{6}x - \frac{36}{3}\), \(y = 1.\dot{6}x - 12\), \(y = \frac{5x}{3} - 12\). Do not allow \(y = 1.67x - 12\). ISW after correct equation found |
## Question 3(a):
Line $l_1$: $3x + 5y - 7 = 0$
| Working/Answer | Marks | Guidance |
|---|---|---|
| Attempts to make $y$ the subject of $3x + 5y - 7 = 0$ | M1 | Expect $\pm 5y = \ldots$ then $y = \ldots$; $\frac{3}{5}$ alone is M0 |
| States $-\frac{3}{5}$ or exact equivalent | A1 | Correct answer implies both marks; value of $c$ need not be correct; do not allow $-\frac{3}{5}x$ alone |
## Question 3(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Uses perpendicular gradients rule $\Rightarrow$ gradient $l_2 = \frac{5}{3}$ | M1 | |
| Forms equation of $l_2$ using $(6, -2)$: $y + 2 = \frac{5}{3}(x - 6)$ | M1 | |
| $y = \frac{5}{3}x - 12$ | A1 | |
## Question (b) [Perpendicular Line]:
| Answer | Mark | Guidance |
|--------|------|----------|
| Uses perpendicular gradient rule following through on gradient from (a) | M1 | If gradient not given, follow through on their $m$ |
| Equation of straight line with **changed** gradient using $(6, -2)$ | M1 | So if (a) was $-\frac{3}{5}$, then $(y+2) = \frac{3}{5}(x-6)$ scores this. At least one bracket correct. If $y=mx+c$ used, must find $c$ using gradient and $(6,-2)$ |
| $y = \frac{5}{3}x - 12$ | A1 | Allow exact equivalents e.g. $y = \frac{10}{6}x - \frac{36}{3}$, $y = 1.\dot{6}x - 12$, $y = \frac{5x}{3} - 12$. Do not allow $y = 1.67x - 12$. ISW after correct equation found |
---\begin{enumerate}
\item The line $l _ { 1 }$ has equation $3 x + 5 y - 7 = 0$\\
(a) Find the gradient of $l _ { 1 }$
\end{enumerate}
The line $l _ { 2 }$ is perpendicular to $l _ { 1 }$ and passes through the point $( 6 , - 2 )$.\\
(b) Find the equation of $l _ { 2 }$ in the form $y = m x + c$, where $m$ and $c$ are constants.\\
\hfill \mbox{\textit{Edexcel P1 2019 Q3 [5]}}