Moderate -0.8 This is a straightforward C1 coordinate geometry question requiring standard procedures: find midpoint, calculate gradient of PQ, use perpendicular gradient relationship (m₁m₂ = -1), then write equation in required form. All steps are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial since it requires multiple coordinated steps.
The points \(P\) and \(Q\) have coordinates \(( - 1,6 )\) and \(( 9,0 )\) respectively.
The line \(l\) is perpendicular to \(P Q\) and passes through the mid-point of \(P Q\).
Find an equation for \(l\), giving your answer in the form \(a x + b y + c = 0\), where \(a\), \(b\) and \(c\) are integers.
\(PQ: m = \frac{0-6}{9-(-1)} = \left(-\frac{3}{5}\right)\)
B1
B1: correct numerical expression for gradient – need not be simplified
Gradient perpendicular to \(PQ = -\frac{1}{m} \left(=\frac{5}{3}\right)\)
M1
\(y - 3 = \frac{5}{3}(x - 4)\)
M1
\(5x - 3y - 11 = 0\) or \(3y - 5x + 11 = 0\) or multiples e.g. \(10x - 6y - 22 = 0\)
A1
A1: Requires integer form with an = zero (see examples above)
Notes
B1: correct midpoint. B1: correct numerical expression for gradient – need not be simplified. 1st Negative reciprocal of their numerical value for m. 2nd M: Equation of a line through their (4, 3) with any gradient except 0 or ∞. If the 4 and 3 are the wrong way round the 2nd M mark can still be given if a correct formula is seen, (e.g. \(y - y_1 = m(x - x_1)\)) is seen, otherwise M0. If (4, 3) is substituted into \(y = mx + c\) to find \(c\), the 2nd M mark is for attempting this.
Mid-point of $PQ$ is $(4, 3)$ | B1 |
$PQ: m = \frac{0-6}{9-(-1)} = \left(-\frac{3}{5}\right)$ | B1 | B1: correct numerical expression for gradient – need not be simplified
Gradient perpendicular to $PQ = -\frac{1}{m} \left(=\frac{5}{3}\right)$ | M1 |
$y - 3 = \frac{5}{3}(x - 4)$ | M1 |
$5x - 3y - 11 = 0$ or $3y - 5x + 11 = 0$ or multiples e.g. $10x - 6y - 22 = 0$ | A1 | A1: Requires integer form with an = zero (see examples above)
Notes | | B1: correct midpoint. B1: correct numerical expression for gradient – need not be simplified. 1st Negative reciprocal of their numerical value for m. 2nd M: Equation of a line through their (4, 3) with any gradient except 0 or ∞. If the 4 and 3 are the wrong way round the 2nd M mark can still be given if a correct formula is seen, (e.g. $y - y_1 = m(x - x_1)$) is seen, otherwise M0. If (4, 3) is substituted into $y = mx + c$ to find $c$, the 2nd M mark is for attempting this.
---
The points $P$ and $Q$ have coordinates $( - 1,6 )$ and $( 9,0 )$ respectively.
The line $l$ is perpendicular to $P Q$ and passes through the mid-point of $P Q$.\\
Find an equation for $l$, giving your answer in the form $a x + b y + c = 0$, where $a$, $b$ and $c$ are integers.\\
\hfill \mbox{\textit{Edexcel C1 2011 Q3 [5]}}