| Exam Board | SPS |
|---|---|
| Module | SPS SM (SPS SM) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Perpendicular line through point |
| Difficulty | Standard +0.3 This is a multi-part coordinate geometry question requiring perpendicular gradients, distance formula, and solving a quadratic. While it has multiple steps (11 marks total), each component uses standard AS-level techniques: finding perpendicular line equations, applying distance formula to derive a quadratic, and solving it. The 'show that' in part (b) provides the target equation, reducing problem-solving demand. Slightly above average due to length and algebraic manipulation required, but no novel insights needed. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.10f Distance between points: using position vectors |
\includegraphics{figure_2}
Figure 2 is a sketch showing the line $l_1$ with equation $y = 2x - 1$ and the point $A$ with coordinates $(-2, 3)$.
The line $l_2$ passes through $A$ and is perpendicular to $l_1$
\begin{enumerate}[label=(\alph*)]
\item Find the equation of $l_2$ writing your answer in the form $y = mx + c$, where $m$ and $c$ are constants to be found. [3]
\end{enumerate}
The point $B$ and the point $C$ lie on $l_1$ such that $ABC$ is an isosceles triangle with $AB = AC = 2\sqrt{13}$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the $x$ coordinates of points $B$ and $C$ satisfy the equation
$$5x^2 - 12x - 32 = 0$$ [4]
\end{enumerate}
Given that $B$ lies in the 3rd quadrant
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find, using algebra and showing your working, the coordinates of $B$. [4]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS SM 2020 Q3 [11]}}