Grade 10 Maths · Analytical Geometry

Use parallel and perpendicular gradients

Use the fact that parallel lines have equal gradients and perpendicular lines have gradients that multiply to −1, for example 2 and −½.

past paper questions use it: 15 past paper questions use it
videos: 3 videos

Learn it: use parallel and perpendicular gradients

Pick one and watch it, then try the questions below. Videos load only when you tap play.

Analytical Geometry Grade 10: Perpendicular LinesKevinmathscience

How to Find Equation of Straight Line grade 10 maths: Parallel linesKevinmathscience

Analytical Geometry - Parallel and Perpendicular Lines (Grade 10 to 12 Math, Paper 2)MatheNatic

Now try it on a real exam question

No sign-up. Takes 3 minutes.

Question 4.2 · May/June 2025 · Paper 2 · National (DBE) · 4 marks

In the diagram, M is the centre of the circle having equation

(x - 3)^{2} + y^{2} = 25

E (- 1; 3)

and T are points on the circle.

EC is a tangent to the circle at E and cuts the x-axis at D.

E D = \frac{15}{4}

units. MT is produced to meet the tangent at

C (- 7; p)

4.2 Determine the equation of the tangent EC in the form

y = m x + c

.(4)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: use parallel and perpendicular gradients

Work each one out first, then open the memo and check every tick.

3. Question 2.4 · June 2024 · Paper 2 · Provincial · 3 marks
In the diagram below, $A (- 2; 4)$ , $B (- 2; - 1)$ and $C (p; - 4)$ are the vertices of $Δ A BC$ . $Q$ is the x-intercept of the line $A C$ .
$θ$ is the angle of inclination of $A C$ and $B \hat{A} C = 3 7^{\circ}$ .
2.4 Determine the integral value of $p$ if the points $A$ , $Q$ and $C$ are collinear.(3)
Show the memo (3 ticks)Memo
1. ✓
  METHOD 1: Gradient AC set equal to m (method) $\frac{- 4 - 4}{p + 2} = - 1, 33$
  1 mark · Use parallel and perpendicular gradients
2. ✓
  Simplification $- 8 = - 1, 33 p - 2, 66$ $1, 33 p = 5, 34$
  1 mark · Solve a linear equation
3. ✓
  Answer $p = 4$
  1 mark · Use parallel and perpendicular gradients
4. Question 1.5 · June 2024 · Paper 2 · Provincial · 4 marks
$PQRS$ is a quadrilateral with vertices $P (- 5; 0)$ , $Q (0; t)$ , $R (3; - 2)$ and $S (- 2; - 5)$ .
$P$ is on the x-axis and $Q$ is on the y-axis.
1.5 Calculate the size of $P \hat{Q} R$ .(4)
Show the memo (4 ticks)Memo
1. ✓
  METHOD 1: Substitution for the gradient of QR $m_{QR} = \frac{- 2 - 3}{3 - 0}$
  1 mark · Calculate gradient between two points
2. ✓
  Gradient of QR $m_{QR} = - \frac{5}{3}$
  1 mark · Calculate gradient between two points
3. ✓
  Product of the gradients (method) $m_{PQ} \times m_{QR} = \frac{3}{5} \times (- \frac{5}{3}) = - 1$
  1 mark · Use parallel and perpendicular gradients
4. ✓
  Answer $PQ ⊥ QR$ , so $P \hat{Q} R = 9 0^{\circ}$
  1 mark · Use parallel and perpendicular gradients
5. Question 3.5 · March 2016 · Provincial · 3 marks
3.5 If the following points $B (x; - 3)$ , $E (- 4; 5)$ and $F (2; 2)$ are collinear, calculate the value of x.(3)
Show the memo (3 ticks)Memo
1. ✓
  Equating gradients (M) $m_{BE} = m_{EF}$ : $\frac{5 - ( - 3 )}{- 4 - x} = \frac{2 - 5}{2 - ( - 4 )}$
  1 mark · Use parallel and perpendicular gradients
2. ✓
  Correct gradients $\frac{8}{- 4 - x} = - \frac{1}{2}$
  1 mark · Calculate gradient between two points
3. ✓
  Answer $16 = - 1 (- 4 - x) = 4 + x$ , so $x = 12$
  1 mark · Use parallel and perpendicular gradients
6. Question 5.5 · June 2024 · Paper 1 · Provincial · 3 marks
Sketched below are the graphs of $f (x) = a x^{2} + b x + c$ and $g (x) = m x + k$ . The x-intercepts of f are at points A and C. The graphs of f and g intersect at points A and B respectively. Point P is the turning point of f. Q is a point on g such that line PQ is parallel to the y-axis. Points given: $A (- 4; 0)$ , $C (2; 0)$ , $B (0; 8)$ .
5.5 Determine the equation of a line p perpendicular to g passing through point C.(3)
Show the memo (3 ticks)Memo
1. ✓
  Determine gradient $m_{g} = 2$
  Perpendicular gradient $m_{p} = - \frac{1}{2}$
  1 mark · Use parallel and perpendicular gradients
2. ✓
  Substitute point C(2;0) $y - 0 = - \frac{1}{2} (x - 2)$
  1 mark · Find the equation of a line
3. ✓
  Standard form $y = - \frac{1}{2} x + 1$
  1 mark · Find the equation of a line

Free on Android

Know exactly which skill is costing you marks.

Mark past papers tick by tick in the NavyBlue app. It remembers every mark you lose, shows you the skills to fix first, and picks your next questions.

The NavyBlue app showing a learner's weak skills