| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Equation of line through two points |
| Difficulty | Standard +0.3 This is a standard C1 coordinate geometry question requiring gradient calculation, perpendicular line equations, and distance verification. While it has multiple parts and requires careful algebraic manipulation, all techniques are routine textbook exercises with no novel problem-solving insight needed. Slightly above average difficulty due to the multi-step nature and the need to verify the isosceles property using distance formula. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\text{grad} = \frac{1-5}{4-(-2)} = -\frac{2}{3}\) | M1 A1 | |
| \(y - 5 = -\frac{2}{3}(x + 2)\) | M1 | |
| \(3y - 15 = -2x - 4\), so \(2x + 3y = 11\) | A1 | |
| (b) \(\text{grad } l_2 = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}\) | M1 A1 | |
| \(y - 1 = \frac{3}{2}(x - 4)\) (or \([3x - 2y = 10]\)) | A1 | |
| (c) At \(C\), \(x = 0\): \(y = -5\), so \(C(0, -5)\) | B1 | |
| \(AB = \sqrt{(4+2)^2 + (1-5)^2} = \sqrt{36 + 16} = \sqrt{52}\) | M1 A1 | |
| \(BC = \sqrt{(0-4)^2 + (-5-1)^2} = \sqrt{16 + 36} = \sqrt{52}\) | ||
| \(AB = BC\), therefore triangle \(ABC\) is isosceles | A1 | (11) |
**(a)** $\text{grad} = \frac{1-5}{4-(-2)} = -\frac{2}{3}$ | M1 A1 |
$y - 5 = -\frac{2}{3}(x + 2)$ | M1 |
$3y - 15 = -2x - 4$, so $2x + 3y = 11$ | A1 |
**(b)** $\text{grad } l_2 = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}$ | M1 A1 |
$y - 1 = \frac{3}{2}(x - 4)$ (or $[3x - 2y = 10]$) | A1 |
**(c)** At $C$, $x = 0$: $y = -5$, so $C(0, -5)$ | B1 |
$AB = \sqrt{(4+2)^2 + (1-5)^2} = \sqrt{36 + 16} = \sqrt{52}$ | M1 A1 |
$BC = \sqrt{(0-4)^2 + (-5-1)^2} = \sqrt{16 + 36} = \sqrt{52}$ | |
$AB = BC$, therefore triangle $ABC$ is isosceles | A1 | (11)8. The straight line $l _ { 1 }$ passes through the point $A ( - 2,5 )$ and the point $B ( 4,1 )$.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for $l _ { 1 }$ in the form $a x + b y = c$, where $a , b$ and $c$ are integers.
The straight line $l _ { 2 }$ passes through $B$ and is perpendicular to $l _ { 1 }$.
\item Find an equation for $l _ { 2 }$.
Given that $l _ { 2 }$ meets the $y$-axis at the point $C$,
\item show that triangle $A B C$ is isosceles.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q8 [11]}}