Edexcel C1 — Question 3 5 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStraight Lines & Coordinate Geometry
TypePerpendicular line through point
DifficultyModerate -0.8 This is a straightforward C1 question testing perpendicular gradients and point-slope form. Students need to rearrange to find the gradient of l (1/5), apply the perpendicular gradient rule (-5), then substitute the point to find c. All steps are routine procedures with no problem-solving required, making it easier than average but worth 5 marks due to multiple computational steps.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships
The straight line \(l\) has the equation \(x - 5y = 7\). The straight line \(m\) is perpendicular to \(l\) and passes through the point \((-4, 1)\). Find an equation for \(m\) in the form \(y = mx + c\). [5]
AnswerMarks Guidance
\(x - 5y = 7 \Rightarrow y = \frac{1}{5}x - \frac{7}{5} \therefore \text{grad} = \frac{1}{5}\)B1
\(\text{grad } m = \frac{-1}{\frac{1}{5}} = -5\)M1 A1
\(\therefore y - 1 = -5(x + 4)\)M1
\(y = -5x - 19\)A1 (5) 
| $x - 5y = 7 \Rightarrow y = \frac{1}{5}x - \frac{7}{5} \therefore \text{grad} = \frac{1}{5}$ | B1 |
| $\text{grad } m = \frac{-1}{\frac{1}{5}} = -5$ | M1 A1 |
| $\therefore y - 1 = -5(x + 4)$ | M1 |
| $y = -5x - 19$ | A1 | (5)
The straight line $l$ has the equation $x - 5y = 7$.

The straight line $m$ is perpendicular to $l$ and passes through the point $(-4, 1)$.

Find an equation for $m$ in the form $y = mx + c$. [5]

\hfill \mbox{\textit{Edexcel C1  Q3 [5]}}
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