GRADE 11 · Analytical Geometry · Past Question Papers
1 Summary Notes 2 Past Question Papers 3 Test Your Knowledge
Grade 11 · Paper 2 · CAPS Aligned

Analytical Geometry
Past Question Papers

17 questions arranged by DBE cognitive level — the equation of a line through two points, through one point parallel or perpendicular to a given line, and the inclination of a line, finishing with a perpendicular bisector and a median. Work each one on paper first, then reveal the memo.

17
practice questions
4
cognitive levels
17
worked memos
100%
independently verified
How to use this bank.
  1. Start at Level 1 and move up — don't jump to Level 4 first.
  2. Always calculate the gradient FIRST (Grade 10's formula) before trying to find any equation.
  3. Rearrange any general-form equation (\(Ax+By+C=0\)) into \(y=mx+c\) form before reading off its gradient.
  4. Reveal the memo only after a genuine attempt.
Accuracy note: every question below was independently solved from scratch in Python before publication, cross-checked against its own working rather than assumed correct. All 17 are original "Equation Station SA Practice Question" items — a real DBE/provincial exam-archive citation pass for Grade 11 Analytical Geometry has not yet been run.
L1 — Knowledge 4 Qs
L2 — Routine Procedures 5 Qs
L3 — Complex Procedures 4 Qs
L4 — Problem Solving 4 Qs
24%
Level 1 | Knowledge
Direct Substitution

One direct application of the equation-of-a-line method, plus recognising inclination and the parallel condition.

Q1Equation Station SA Practice Question2 marks
Equation From Gradient + Point
Equation From a Gradient and a Point

Find the equation of the line with gradient 3, passing through \((2,4)\).

Memo
✓ \(y=3x+c\). Substitute \((2,4)\): \(4=3(2)+c\Rightarrow c=-2\)✓ \(\boxed{y=3x-2}\)
Q2Equation Station SA Practice Question2 marks
Equation From Two Points
Equation From Two Points

Find the equation of the line through \((0,0)\) and \((2,10)\).

Memo
✓ \(m=\dfrac{10-0}{2-0}=5\). Substitute \((0,0)\): \(0=5(0)+c\Rightarrow c=0\)✓ \(\boxed{y=5x}\)
Q3Equation Station SA Practice Question1 mark
Inclination
Gradient to Inclination

Calculate the inclination of a line with gradient 1.

Memo
✓ \(\tan\theta=1\Rightarrow\boxed{\theta=45°}\)
Q4Equation Station SA Practice Question1 mark
Parallel Lines
Recognising the Parallel Gradient

A line is parallel to \(y=5x-2\). What is its gradient?

Memo
✓ A parallel line has the same gradient: \(\boxed{m=5}\)
29%
Level 2 | Routine Procedures
One Established Method

Negative gradients, finding a perpendicular gradient, correcting a calculator's inclination answer, and rearranging general form.

Q5Equation Station SA Practice Question2 marks
Equation From Gradient + Point
Negative Gradient

Find the equation of the line with gradient \(-2\), passing through \((3,-1)\).

Memo
✓ \(y=-2x+c\). Substitute \((3,-1)\): \(-1=-2(3)+c=-6+c\Rightarrow c=5\)✓ \(\boxed{y=-2x+5}\)
Q6Equation Station SA Practice Question2 marks
Equation From Two Points
Two Points, Negative Gradient

Find the equation of the line through \((-2,8)\) and \((2,0)\).

Memo
✓ \(m=\dfrac{0-8}{2-(-2)}=\dfrac{-8}{4}=-2\). Substitute \((2,0)\): \(0=-2(2)+c\Rightarrow c=4\)✓ \(\boxed{y=-2x+4}\)
Q7Equation Station SA Practice Question2 marks
Perpendicular Lines
Finding a Perpendicular Gradient

A line has a gradient of 4. Calculate the gradient of a line perpendicular to it.

Memo
✓ \(m_{\perp}=-\dfrac{1}{4}\Rightarrow\boxed{-\dfrac{1}{4}}\)
Q8Equation Station SA Practice Question2 marks
Inclination
Negative Gradient to Inclination

Calculate the inclination of a line with gradient \(-1\).

Memo
✓ \(\tan\theta=-1\) gives a raw calculator value of \(-45°\); since inclination must be between \(0°\) and \(180°\), add \(180°\)✓ \(-45°+180°=\boxed{135°}\)
Q9Equation Station SA Practice Question2 marks
General Form
Reading the Gradient From General Form

Calculate the gradient of the line \(3x+4y-8=0\).

Memo
✓ Rearrange into \(y=mx+c\) form: \(4y=-3x+8\Rightarrow y=-\dfrac{3}{4}x+2\)✓ \(\boxed{m=-\dfrac{3}{4}}\)
24%
Level 3 | Complex Procedures
Multi-Step Methods

Working backward for an unknown coordinate, parallel/perpendicular lines starting from general form, and inclination working backward.

Q10Equation Station SA Practice Question3 marks
Equation From Two Points
Finding an Unknown Coordinate

The line through \((2,3)\) and \((5,k)\) has equation \(y=4x-5\). Calculate \(k\).

Memo
✓ Since \((5,k)\) lies on the line, substitute \(x=5\): \(k=4(5)-5\)✓ \(\boxed{k=15}\)
Q11Equation Station SA Practice Question4 marks
Parallel Lines
Parallel Line From General Form

Find the equation of the line through \((4,-2)\), parallel to \(2x-y+1=0\).

Memo
✓ Rearrange first: \(y=2x+1\), so \(m=2\)✓ Substitute \((4,-2)\): \(-2=2(4)+c=8+c\Rightarrow c=-10\)✓ \(\boxed{y=2x-10}\)
Q12Equation Station SA Practice Question4 marks
Perpendicular Lines
Perpendicular Line From General Form

Find the equation of the line through \((1,4)\), perpendicular to \(x+2y-6=0\).

Memo
✓ Rearrange first: \(y=-\dfrac{1}{2}x+3\), so \(m=-\dfrac{1}{2}\)✓ Perpendicular gradient: \(m_{\perp}=-\dfrac{1}{-1/2}=2\)✓ Substitute \((1,4)\): \(4=2(1)+c\Rightarrow c=2\)✓ \(\boxed{y=2x+2}\)
Q13Equation Station SA Practice Question2 marks
Inclination
Inclination to Gradient

A line has an inclination of \(120°\). Calculate its gradient.

Memo
✓ \(m=\tan120°=\boxed{-\sqrt3\approx-1{,}73}\)
24%
Level 4 | Problem Solving
Combined Skills

Solving for an unknown constant, and combining formulas for a perpendicular bisector and a median.

Q14Equation Station SA Practice Question4 marks
Parallel Lines
Solving for an Unknown Constant (Parallel)

Line 1: \(y=2x+1\). Line 2: \(kx-4y+8=0\) is PARALLEL to Line 1. Calculate \(k\).

Memo
✓ Rearrange Line 2: \(4y=kx+8\Rightarrow y=\dfrac{k}{4}x+2\), gradient \(=\dfrac{k}{4}\)✓ Parallel: \(\dfrac{k}{4}=2\Rightarrow\boxed{k=8}\)
Q15Equation Station SA Practice Question4 marks
Perpendicular Lines
Solving for an Unknown Constant (Perpendicular)

Line 1: \(y=0{,}5x-3\). Line 2: \(kx+y-5=0\) is PERPENDICULAR to Line 1. Calculate \(k\).

Memo
✓ Rearrange Line 2: \(y=-kx+5\), gradient \(=-k\)✓ Perpendicular: \(0{,}5\times(-k)=-1\Rightarrow\boxed{k=2}\)
Q16Equation Station SA Practice Question6 marks
Combined
The Perpendicular Bisector

\(A(0,2)\) and \(B(6,10)\). Find the equation of the perpendicular bisector of \(AB\).

Memo
✓ Midpoint of \(AB\): \(M=\left(\dfrac{0+6}{2},\dfrac{2+10}{2}\right)=(3,6)\)✓ Gradient of \(AB\): \(m_{AB}=\dfrac{10-2}{6-0}=\dfrac{4}{3}\)✓ Perpendicular gradient: \(m_{\perp}=-\dfrac{3}{4}\)✓ Substitute \(M(3,6)\): \(6=-\dfrac{3}{4}(3)+c=-\dfrac{9}{4}+c\Rightarrow c=\dfrac{33}{4}\)✓ \(\boxed{y=-\dfrac{3}{4}x+\dfrac{33}{4}}\)
Q17Equation Station SA Practice Question6 marks
Combined
The Median of a Triangle

\(\triangle ABC\) has vertices \(A(5,12)\), \(B(-1,4)\) and \(C(9,-2)\). Find the equation of the median from \(A\) to the midpoint of \(BC\).

Memo
✓ Midpoint of \(BC\): \(D=\left(\dfrac{-1+9}{2},\dfrac{4+(-2)}{2}\right)=(4,1)\)✓ Gradient of \(AD\): \(m_{AD}=\dfrac{1-12}{4-5}=\dfrac{-11}{-1}=11\)✓ Substitute \(A(5,12)\): \(12=11(5)+c=55+c\Rightarrow c=-43\)✓ \(\boxed{y=11x-43}\)
This bank pairs with the Summary Notes tab's worked-example sequence. Ready to check your recall under exam-style pressure? Try the Test Your Knowledge quiz.