17 questions arranged by DBE cognitive level — the distance, gradient and midpoint formulas, and the conditions for parallel and perpendicular lines, finishing with combined shape-classification problems. 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.
Start at Level 1 and move up — don't jump to Level 4 first.
Sketch the points roughly on a grid before calculating anything — a rough sketch catches an obviously wrong final answer immediately.
Label which point is 1 and which is 2 before substituting, and stay consistent within the same calculation.
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 10 Analytical Geometry has not yet been run, unlike the site's other Grade 10 topics.
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 distance, gradient or midpoint formula, plus recognising the parallel condition.
Q1Equation Station SA Practice Question2 marks
Distance
Distance Between Two Points
Calculate the distance between \(A(2,3)\) and \(B(5,7)\).
Memo
✓ Substitute directly into the distance formula: \(d=\sqrt{(5-2)^2+(7-3)^2}=\sqrt{9+16}\)✓ \(\boxed{d=5}\)
Q2Equation Station SA Practice Question2 marks
Gradient
Gradient of a Line Segment
Calculate the gradient of the line through \(A(1,2)\) and \(B(4,11)\).
Memo
✓ Substitute directly into the gradient formula: \(m=\dfrac{11-2}{4-1}=\dfrac{9}{3}\)✓ \(\boxed{m=3}\)
Q3Equation Station SA Practice Question2 marks
Midpoint
Midpoint of a Line Segment
Calculate the midpoint of the line segment joining \(A(2,6)\) and \(B(8,10)\).
Memo
✓ Average the two \(x\)-values and the two \(y\)-values separately: \(M=\left(\dfrac{2+8}{2},\dfrac{6+10}{2}\right)\)✓ \(\boxed{M(5,8)}\)
Q4Equation Station SA Practice Question1 mark
Parallel Lines
Recognising Parallel Lines
Line 1 has a gradient of \(\dfrac{2}{3}\). Line 2 also has a gradient of \(\dfrac{2}{3}\). What can you conclude about the two lines?
Memo
✓ Two lines with exactly the same gradient rise (or fall) at exactly the same rate, so they never meet — that is the definition of parallel.✓ \(\boxed{\text{The lines are parallel}}\)
29%
Level 2 | Routine Procedures
One Established Method
The same three formulas with negative coordinates, plus finding a perpendicular gradient and checking whether two lines are parallel.
Q5Equation Station SA Practice Question2 marks
Distance
Distance With Negative Coordinates
Calculate the distance between \(A(-2,3)\) and \(B(4,-5)\).
Memo
✓ Subtracting a negative coordinate is the same as adding — keep the sign carefully: \(d=\sqrt{(4-(-2))^2+(-5-3)^2}=\sqrt{6^2+(-8)^2}=\sqrt{36+64}\)✓ \(\boxed{d=10}\)
Q6Equation Station SA Practice Question2 marks
Gradient
Gradient With Negative Coordinates
Calculate the gradient of the line through \(A(-1,4)\) and \(B(3,-8)\).
Memo
✓ Substitute carefully, keeping every sign: \(m=\dfrac{-8-4}{3-(-1)}=\dfrac{-12}{4}\)✓ \(\boxed{m=-3}\)
Q7Equation Station SA Practice Question2 marks
Midpoint
Midpoint With Negative Coordinates
Calculate the midpoint of the line segment joining \(A(-4,6)\) and \(B(8,-2)\).
Memo
✓ Average the two \(x\)-values and the two \(y\)-values, tracking the negative signs: \(M=\left(\dfrac{-4+8}{2},\dfrac{6+(-2)}{2}\right)\)✓ \(\boxed{M(2,2)}\)
Q8Equation Station SA Practice Question2 marks
Perpendicular Lines
Finding a Perpendicular Gradient
A line has a gradient of \(\dfrac{4}{5}\). Calculate the gradient of a line perpendicular to it.
Memo
✓ Perpendicular gradients are negative reciprocals of each other — flip the fraction AND change the sign: \(m_2=\dfrac{-1}{4/5}\)✓ \(\boxed{m_2=-\dfrac{5}{4}}\)
Q9Equation Station SA Practice Question3 marks
Parallel Lines
Checking Whether Two Lines Are Parallel
Line 1 passes through \((0,2)\) and \((3,8)\). Line 2 passes through \((1,-1)\) and \((4,5)\). Are the two lines parallel?
Memo
✓ Calculate each line's own gradient separately: \(m_1=\dfrac{8-2}{3-0}=2\)✓ \(m_2=\dfrac{5-(-1)}{4-1}=\dfrac{6}{3}=2\)✓ \(\boxed{m_1=m_2=2}\), so the lines are parallel
24%
Level 3 | Complex Procedures
Working Backward
Using a known distance, gradient, or midpoint to find a missing coordinate — plus proving three points are collinear.
Q10Equation Station SA Practice Question4 marks
Distance
Finding an Unknown Coordinate From a Distance
\(A(3,2)\) and \(B(x,10)\) are 10 units apart. Calculate the possible value(s) of \(x\).
Memo
✓ Substitute into the distance formula and square both sides to remove the square root: \(10=\sqrt{(x-3)^2+8^2}\Rightarrow100=(x-3)^2+64\)✓ Isolate the squared term, then take the square root of both sides — remembering there are two possible roots: \((x-3)^2=36\Rightarrow x-3=\pm6\)✓ \(\boxed{x=9\text{ or }x=-3}\)
Q11Equation Station SA Practice Question3 marks
Gradient
Finding an Unknown Coordinate From a Gradient
The line through \(A(2,-3)\) and \(B(6,y)\) has a gradient of 1,5. Calculate \(y\).
Memo
✓ Substitute into the gradient formula and solve for \(y\): \(1{,}5=\dfrac{y-(-3)}{6-2}\Rightarrow1{,}5\times4=y+3\)✓ \(6=y+3\Rightarrow\boxed{y=3}\)
Q12Equation Station SA Practice Question4 marks
Midpoint
Finding a Missing Endpoint
\(M(4,-1)\) is the midpoint of \(A(-2,5)\) and \(B\). Calculate the coordinates of \(B\).
Memo
✓ Substitute into the midpoint formula for each coordinate separately: \(4=\dfrac{-2+x_B}{2}\Rightarrow x_B=8-(-2)=10\)✓ \(-1=\dfrac{5+y_B}{2}\Rightarrow y_B=-2-5=-7\)✓ \(\boxed{B(10,-7)}\)
Q13Equation Station SA Practice Question3 marks
Gradient
Proving Three Points Are Collinear
Show that \(A(1,1)\), \(B(3,5)\) and \(C(6,11)\) lie on the same straight line.
Memo
✓ Calculate the gradient between the first pair of points: \(m_{AB}=\dfrac{5-1}{3-1}=\dfrac{4}{2}=2\)✓ Calculate the gradient between the second pair, sharing point \(B\): \(m_{BC}=\dfrac{11-5}{6-3}=\dfrac{6}{3}=2\)✓ \(\boxed{m_{AB}=m_{BC}=2}\), and both pass through \(B\), so \(A\), \(B\) and \(C\) are collinear
24%
Level 4 | Problem Solving
Combined Skills
Classifying triangles and quadrilaterals by combining distance and gradient in a single proof.
Q14Equation Station SA Practice Question5 marks
Combined
Right-Angled and Isosceles
\(P(1,1)\), \(Q(4,5)\) and \(R(8,2)\) are the vertices of \(\triangle PQR\). Show that \(\triangle PQR\) is both right-angled at \(Q\) and isosceles.
Memo
✓ For the right angle, multiply the gradients of the two sides that meet at \(Q\): \(m_{PQ}=\dfrac{5-1}{4-1}=\dfrac{4}{3}\), \(m_{QR}=\dfrac{2-5}{8-4}=-\dfrac{3}{4}\)✓ \(m_{PQ}\times m_{QR}=\dfrac{4}{3}\times\left(-\dfrac{3}{4}\right)=-1\), so \(\triangle PQR\) is right-angled at \(Q\)✓ For isosceles, calculate the two sides meeting at \(Q\): \(PQ=\sqrt{3^2+4^2}=5\), \(QR=\sqrt{4^2+3^2}=5\)✓ \(\boxed{PQ=QR=5}\), so \(\triangle PQR\) is right-angled at \(Q\) and isosceles — you never needed \(PR\) at all
Q15Equation Station SA Practice Question4 marks
Combined
Proving a Parallelogram via Diagonals
\(ABCD\) has vertices \(A(1,2)\), \(B(6,3)\), \(C(8,7)\) and \(D(3,6)\). Show that \(ABCD\) is a parallelogram.
Memo
✓ A quadrilateral is a parallelogram if its diagonals bisect each other — calculate the midpoint of each diagonal: midpoint of \(AC\): \(\left(\dfrac{1+8}{2},\dfrac{2+7}{2}\right)=(4{,}5;\ 4{,}5)\)✓ midpoint of \(BD\): \(\left(\dfrac{6+3}{2},\dfrac{3+6}{2}\right)=(4{,}5;\ 4{,}5)\)✓ \(\boxed{\text{Both diagonals share the same midpoint }(4{,}5;\ 4{,}5)}\), so they bisect each other and \(ABCD\) is a parallelogram
Q16Equation Station SA Practice Question6 marks
Combined
Proving a Square
\(ABCD\) has vertices \(A(0,0)\), \(B(4,3)\), \(C(1,7)\) and \(D(-3,4)\). Prove that \(ABCD\) is a square.
Memo
✓ Calculate all four sides using the distance formula: \(AB=\sqrt{4^2+3^2}=5\), \(BC=\sqrt{3^2+4^2}=5\), \(CD=\sqrt{4^2+3^2}=5\), \(DA=\sqrt{3^2+4^2}=5\)✓ \(\boxed{AB=BC=CD=DA=5}\) — all four sides equal, so \(ABCD\) is at least a rhombus✓ Now check one angle using gradient: \(m_{AB}=\dfrac{3}{4}\), \(m_{BC}=\dfrac{7-3}{1-4}=-\dfrac{4}{3}\)✓ \(m_{AB}\times m_{BC}=\dfrac{3}{4}\times\left(-\dfrac{4}{3}\right)=-1\), so the angle at \(B\) is \(90°\) — a rhombus with one right angle is a square, so \(\boxed{ABCD\text{ is a square}}\)
Q17Equation Station SA Practice Question5 marks
Combined
Finding the Fourth Vertex of a Parallelogram
\(KLMN\) is a parallelogram with \(K(1,2)\), \(L(5,4)\) and \(M(7,9)\). Calculate the coordinates of \(N\).
Memo
✓ In parallelogram \(KLMN\), the diagonals are \(KM\) and \(LN\) — they must bisect each other, so both share the same midpoint. Calculate the midpoint of \(KM\): \(\left(\dfrac{1+7}{2},\dfrac{2+9}{2}\right)=(4;\ 5{,}5)\)✓ This same point must also be the midpoint of \(LN\) — set up the midpoint formula for \(LN\) and solve for \(N\)'s coordinates: \(4=\dfrac{5+x_N}{2}\Rightarrow x_N=8-5=3\)✓ \(5{,}5=\dfrac{4+y_N}{2}\Rightarrow y_N=11-4=7\)✓ \(\boxed{N(3,7)}\)