Grade 11 Maths · Analytical Geometry · Find a median, altitude or bisector

Find a median, altitude or bisector Grade 11

Write the equation of a median (through a vertex and the mid-point of the opposite side), an altitude (perpendicular to a side through the opposite vertex) or a perpendicular bisector of a side of a triangle.

questions use it, all grades · 4 past paper, 15 practice: 19 questions use it, all grades · 4 past paper, 15 practice
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How to determine the equation of a median in a triangleKevinmathscience

How to determine the equation of an altitude in a triangleKevinmathscience

Lesson 4: How To Find the Equation of a Perpendicular Bisector | Analytical GeometryMlungisi Nkosi Maths & Science

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Question 4.3 · May/June 2024 · Paper 2 · National (DBE) · 4 marks

In the diagram, the circle centred at

C (2; p)

is drawn.

A (5; 1)

and

B (- 3; - 3)

are points on the circle. E is the midpoint of AB. $Diagram for this question$

4.3 Determine the equation of the perpendicular bisector of AB 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 questions: find a median, altitude or bisector

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3. Question 3.2 · September 2026 · Paper 2 · Provincial · 3 marks
In the diagram A $(- 2; - 3)$ , B $(3; - 4)$ , C $(4; r)$ and D $(2; 1)$ are the vertices of quadrilateral ABCD. $Diagram for this question$
3.2 Determine the equation of BM, the median of $△$ ABD.(3)
Show the memo (3 ticks)Memo
1. ✓
  Gradient of BM $m_{BM} = \frac{- 4 - ( - 1 )}{3 - 0} = - 1$
  1 mark · Find a median, altitude or bisector
2. ✓
  Substitution in formula $y - y_{1} = m (x - x_{1})$
  $y - (- 4) = - 1 (x - 3)$ or $y - (- 1) = - 1 (x - 0)$
  1 mark · Find the equation of a line
3. ✓
  Answer $y + 4 = - x + 3$
  $y = - x - 1$
  1 mark · Find the equation of a line
4. Question 4.3 · November 2024 · Paper 2 · Provincial · 3 marks
In the diagram below, P(-1 ; 3), R(3 ; -1) and S( $a$ ; $b$ ) are the vertices of $Δ PRS$ .
Point T lies on the perpendicular bisector of PR. $Diagram for this question$
4.3 Determine the equation of ST, if ST $⊥$ PR and $m_{PR} = - 1$ .(3)
Show the memo (3 ticks)Memo
1. ✓
  $m_{ST}$ $m_{PR} \times m_{2} = - 1$
  $(- 1) \times (m_{2}) = - 1$
  $∴ m_{ST} = 1$
  1 mark · Use parallel and perpendicular gradients
2. ✓
  Substitute (1 ; 1) $y - y_{1} = m (x - x_{1})$
  $y - (1) = 1 (x - 1)$
  1 mark · Find a median, altitude or bisector
3. ✓
  Equation $y - 1 = x - 1$
  $y = x$
  1 mark · Find the equation of a line
5. Practice · NavyBlue · 2 marks
$△ A BC$ has vertices $A (3; 7)$ , $B (- 1; - 1)$ and $C (7; 1)$ .
(a) Determine the equation of the median from $A$ .(2)
Show the memo (2 ticks)Memo
1. ✓
  Mid-point of BC $(\frac{- 1 + 7}{2}; \frac{- 1 + 1}{2}) = (3; 0)$
  1 mark · Calculate the mid-point of a segment
2. ✓
  Equation of the median $A (3; 7)$ and $(3; 0)$ have the same $x$ -value, so the median is $x = 3$
  1 mark · Find a median, altitude or bisector
6. Practice · NavyBlue · 2 marks
$A BC D$ is a quadrilateral with vertices $A (- 2; 0)$ , $B (1; 3)$ , $C (6; 0)$ and $D (1; - 3)$ .
(b) Show that the diagonal $A C$ is the perpendicular bisector of the diagonal $B D$ .(2)
Show the memo (2 ticks)Memo
1. ✓
  Mid-point of BD $(\frac{1 + 1}{2}; \frac{3 - 3}{2}) = (1; 0)$ , which lies on $A C$ (the $x$ -axis)
  1 mark · Calculate the mid-point of a segment
2. ✓
  Perpendicular bisector $B D$ is vertical ( $x = 1$ ) and $A C$ is horizontal ( $y = 0$ ), so $A C ⊥ B D$
  $A C$ is the perpendicular bisector of $B D$
  1 mark · Find a median, altitude or bisector

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