17 questions arranged by DBE cognitive level — distance, midpoint, gradient, equations of lines and circles, tangents, and proving quadrilaterals. Work each one on paper first, then reveal the memo.
17
practice questions
4
cognitive levels
P2
Paper 2 focus
100%
independently verified
How to use this bank.
Start at Level 1 and move up — don't jump to Level 4 first.
Sketch a quick diagram for any question involving a circle or a quadrilateral.
Substitute carefully, especially with negative coordinates — a sign error here is the most common way to lose marks.
Reveal the memo only after a genuine attempt.
Accuracy note: every question below was independently solved before publication. Source labels are exact — a specific citation means the question was checked against that real archived paper; anything not verifiable against the archive is labelled "Equation Station SA Practice Question," not a fabricated citation.
L1 — Knowledge 4 Qs
L2 — Routine Procedures 5 Qs
L3 — Complex Procedures 4 Qs
L4 — Problem Solving 4 Qs
20%
Level 1 | Knowledge
Direct Formula Application
One formula, one substitution. If these feel shaky, go back to the "Working With Two Points" slides in the Summary Notes before continuing.
Q1Equation Station SA Practice Question3 marks
Distance
Distance Between Two Points
Determine the length of \(PQ\), given \(P(1,2)\) and \(Q(7,10)\).
Memo
✓ \(PQ=\sqrt{(7-1)^2+(10-2)^2}\)✓ \(=\sqrt{36+64}=\sqrt{100}\)✓ \(PQ=10\) units
Q2Equation Station SA Practice Question2 marks
Midpoint
Midpoint of a Segment
Determine the coordinates of \(M\), the midpoint of \(AB\), given \(A(-4,6)\) and \(B(2,-2)\).
Determine the gradient of \(CD\), given \(C(-3,-1)\) and \(D(5,3)\), and state whether the line rises or falls from left to right.
Memo
✓ \(m_{CD}=\dfrac{3-(-1)}{5-(-3)}=\dfrac{4}{8}=\dfrac12\)✓ Since \(m>0\), the line rises from left to right
Q4KZN Preparatory Exam, Sept 20252 marks
Circle
Reading Off Centre and Radius
The equation of a circle with centre \(M\) is \((x+3)^2+(y-4)^2=26\). Write down (a) the coordinates of \(M\), and (b) the length of the radius, in surd form.
Memo
✓ Compare with \((x-a)^2+(y-b)^2=r^2\): here \(a=-3\) and \(b=4\)✓ (a) \(M(-3,4)\)✓ (b) \(r^2=26\Rightarrow r=\sqrt{26}\) (26 has no square factors, so this is already simplest surd form)
35%
Level 2 | Routine Procedures
Standard Methods, One Extra Step
The bulk of Paper 2's opening analytical geometry marks live here. Fluency with surds and inclination matters more than speed at this stage.
Q5Mpumalanga (MDE) Preparatory Exam, June 20252 marks
Distance
Distance Left as a Surd
In \(\triangle PQR\), \(P(5,2)\), \(Q(1,-1)\) and \(R(9,-5)\). Determine the length of \(QR\).
Memo
✓ \(QR=\sqrt{(9-1)^2+(-5-(-1))^2}=\sqrt{64+16}=\sqrt{80}\)✓ \(\sqrt{80}=\sqrt{16\times5}=4\sqrt5\) units
Q6Equation Station SA Practice Question4 marks
Inclination
Gradient, Then Inclination
Determine the gradient of \(EF\) and the inclination of \(EF\), given \(E(1,1)\) and \(F(-2,7)\).
Memo
✓ \(m_{EF}=\dfrac{7-1}{-2-1}=\dfrac{6}{-3}=-2\)✓ Since \(m<0\), \(\theta=180^{\circ}-\tan^{-1}(2)\)✓ \(\tan^{-1}(2)\approx63.43^{\circ}\), so \(\theta\approx180^{\circ}-63.43^{\circ}\)✓ \(\theta\approx116.6^{\circ}\) (to one decimal place)
Q7Equation Station SA Practice Question3 marks
Parallel & perpendicular
Parallel and Perpendicular Gradients
Line \(AB\) has gradient \(-\dfrac35\). Determine (a) the gradient of a line parallel to \(AB\), and (b) the gradient of a line perpendicular to \(AB\).
Memo
✓ (a) A parallel line has the same gradient: \(-\dfrac35\)✓ (b) Perpendicular: \(-\dfrac35\times m_2=-1\Rightarrow m_2=\dfrac53\)
Q8Equation Station SA Practice Question3 marks
Circle
Circle Centred at the Origin
A circle with centre \(O(0,0)\) passes through the point \(P(-6,8)\). Determine the equation of the circle.
Determine the equation of the line through \(K(-1,4)\) and \(L(3,-4)\).
Memo
✓ \(m_{KL}=\dfrac{-4-4}{3-(-1)}=\dfrac{-8}{4}=-2\)✓ \(y-4=-2(x-(-1))=-2x-2\)✓ \(y=-2x-2+4=-2x+2\)✓ Check with \(L(3,-4)\): \(-2(3)+2=-4\) ✓
30%
Level 3 | Complex Procedures
Multi-Step Methods
Collinearity, tangents, and completing the square — no single formula solves these in one line.
Q10Equation Station SA Practice Question4 marks
Collinearity
Proving Three Points Are Collinear
Show that \(P(-5,-3)\), \(Q(-1,0)\) and \(R(7,6)\) are collinear.
Memo
✓ \(m_{PQ}=\dfrac{0-(-3)}{-1-(-5)}=\dfrac{3}{4}\)✓ \(m_{QR}=\dfrac{6-0}{7-(-1)}=\dfrac{6}{8}=\dfrac34\)✓ Since \(m_{PQ}=m_{QR}=\dfrac34\) and \(Q\) is common to both, \(P\), \(Q\) and \(R\) are collinear
Q11Equation Station SA Practice Question6 marks
Circle
Circle Equation, Then Test a Point
A circle has centre \(C(-1,2)\) and passes through \(D(3,5)\). (a) Determine the equation of the circle. (b) Determine whether \(E(-4,-2)\) lies on, inside, or outside the circle.
Memo
✓ (a) \(r^2=(3-(-1))^2+(5-2)^2=16+9=25\)✓ Equation: \((x+1)^2+(y-2)^2=25\)✓ (b) \(CE^2=(-4-(-1))^2+(-2-2)^2=(-3)^2+(-4)^2=9+16=25\)✓ Since \(CE^2=25=r^2\), \(E\) lies exactly ON the circle
Q12Equation Station SA Practice Question5 marks
Tangent
Equation of a Tangent to a Circle
A circle has centre \(C(1,3)\) and passes through \(D(4,7)\). Determine the equation of the tangent to the circle at \(D\).
The equation of a circle is given by \(x^2+y^2-4x+6y-12=0\). Determine the coordinates of the centre and the length of the radius.
Memo
✓ Group: \((x^2-4x)+(y^2+6y)=12\)✓ Complete the square: \((x-2)^2-4+(y+3)^2-9=12\)✓ \((x-2)^2+(y+3)^2=12+4+9=25\)✓ Centre \((2,-3)\), radius \(\sqrt{25}=5\)
15%
Level 4 | Problem Solving
Full Riders, Multiple Skills Combined
Quadrilateral proofs and applied problems that chain two or three formulas together — plan your route before you start substituting.
Q14Equation Station SA Practice Question5 marks
Quadrilaterals
Proving a Parallelogram
Quadrilateral \(ABCD\) has vertices \(A(-2,-1)\), \(B(4,1)\), \(C(6,7)\) and \(D(0,5)\). Prove that \(ABCD\) is a parallelogram.
Memo
✓ A parallelogram's diagonals bisect each other — find the midpoint of \(AC\) and of \(BD\)✓ Midpoint \(AC\): \(\left(\dfrac{-2+6}{2},\dfrac{-1+7}{2}\right)=(2,3)\)✓ Midpoint \(BD\): \(\left(\dfrac{4+0}{2},\dfrac{1+5}{2}\right)=(2,3)\)✓ Both diagonals share the midpoint \((2,3)\), so \(ABCD\) is a parallelogram
Q15Equation Station SA Practice Question5 marks
Applications
Proving a Right Angle in a Triangle
Triangle \(PQR\) has vertices \(P(-3,2)\), \(Q(1,5)\) and \(R(4,1)\). Prove that \(\hat{Q}=90^{\circ}\).
Memo
✓ \(m_{PQ}=\dfrac{5-2}{1-(-3)}=\dfrac34\)✓ \(m_{QR}=\dfrac{1-5}{4-1}=-\dfrac43\)✓ \(m_{PQ}\times m_{QR}=\dfrac34\times\left(-\dfrac43\right)=-1\)✓ Since the product is \(-1\), \(PQ\perp QR\), so \(\hat{Q}=90^{\circ}\)
Q16Equation Station SA Practice Question6 marks
Applications
Equation of a Perpendicular Bisector
Determine the equation of the perpendicular bisector of the line segment joining \(S(-2,3)\) and \(T(6,-1)\).
Quadrilateral \(WXYZ\) has vertices \(W(-4,1)\), \(X(0,4)\), \(Y(4,1)\) and \(Z(0,-2)\). Prove that \(WXYZ\) is a rhombus.
Memo
✓ Midpoint \(WY\): \(\left(\dfrac{-4+4}{2},\dfrac{1+1}{2}\right)=(0,1)\)✓ Midpoint \(XZ\): \(\left(\dfrac{0+0}{2},\dfrac{4+(-2)}{2}\right)=(0,1)\)✓ Equal midpoints ⇒ diagonals bisect each other ⇒ \(WXYZ\) is a parallelogram✓ \(m_{WY}=\dfrac{1-1}{4-(-4)}=0\) (horizontal); \(XZ\) is a vertical line (both points have \(x=0\)), so its gradient is undefined✓ A horizontal line and a vertical line are perpendicular, so the diagonals are perpendicular✓ A parallelogram with perpendicular diagonals is a rhombus — confirmed by equal side lengths: \(WX=XY=YZ=ZW=5\)