Grade 10 Maths · Functions

Find the equation of a line

Write the equation of a straight line from its gradient and a point, for example m = 2 through (1; 3) gives y = 2x + 1.

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How to Find Equation of Straight Line grade 10 maths: IntroductionKevinmathscience

Straight Line Grade 10: y = mx+cKevinmathscience

Analytical Geometry - Equation of a line - Grade 11 & 12 Mathematics (Paper 2)MatheNatic

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Question 6.3 · May/June 2025 · Paper 1 · National (DBE) · 4 marks

The graphs of

f (x) = p^{x} + q

and

g (x) = m x + c

are drawn below.

A (3; 4)

is the point of intersection of

f

and

g

B (0; - 3)

is the y-intercept of

f

. E is the x-intercept of

g

6.3 The graph of

g^{- 1}

, the inverse of

g

, also passes through B. Determine the equation of

g

in the form

y = \dots

(4)

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

More past paper questions: find the equation of a line

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3. Question 6.3 · May/June 2024 · Paper 1 · National (DBE) · 2 marks
The graphs of $f (x) = x^{2} - 2 x - 3$ and $g (x) = m x + c$ are drawn below. D and E are the $x$ -intercepts and P is the $y$ -intercept of $f$ . The turning point of $f$ is T $(1; - 4)$ . The graphs of $f$ and $g$ intersect at P and E.
6.3 Determine the equation of g.(2)
Show the memo (2 ticks)Memo
1. ✓
  Calculate gradient using points P(0;-3) and E(3;0) $m = \frac{0 - ( - 3 )}{3 - 0} = 1$
  1 mark · Calculate gradient between two points
2. ✓
  State final linear equation $y = x - 3$
  1 mark · Find the equation of a line
4. Question 3.4 · May/June 2025 · Paper 2 · National (DBE) · 5 marks
In the diagram below, $Δ SRT$ is drawn where R lies on the x-axis and S lies to the left of R. T lies on the y-axis and the coordinates of S are $(m; 1)$ .

The equation of RT is $2 x - y + 10 = 0$ .
3.4 It is further given that V lies on ST such that VR is perpendicular to ST.

Determine the equation of VR in the form $y = m x + c$ .(5)
Show the memo (3 ticks)Memo
1. ✓
  Calculate the gradient of ST $m_{ST} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
  $m_{ST} = \frac{1 - 10}{- 12 - 0}$
  $m_{ST} = \frac{- 9}{- 12} = \frac{3}{4}$
  2 marks · Calculate gradient between two points
2. ✓
  Determine the gradient of VR $m_{ST} \times m_{V R} = - 1$
  $\frac{3}{4} \times m_{V R} = - 1$
  $m_{V R} = - \frac{4}{3}$
  1 mark · Use parallel and perpendicular gradients
3. ✓
  Find the equation of VRUsing point $R (- 5; 0)$ :
  $y - y_{1} = m (x - x_{1})$
  $y - 0 = - \frac{4}{3} (x - (- 5))$
  $y = - \frac{4}{3} (x + 5)$
  $y = - \frac{4}{3} x - \frac{20}{3}$
  2 marks · Find the equation of a line
5. Question 6.2 · November 2024 · Paper 1 · National (DBE) · 3 marks
In the diagram below, the graphs of $f (x) = - x^{2} + 4 x + 5$ and $g$ , a straight line, are drawn. $C (3; 8)$ is a point of intersection of $f$ and $g$ . $E H$ is drawn parallel to the y-axis, with $E$ a point on $f$ and $H$ a point on $g$ .
6.2 Show that the equation of the line through $A$ and $C$ is given by $g (x) = 2 x + 2$ .(3)
Show the memo (3 ticks)Memo
1. ✓
  Determine coordinates of A (x-intercept) $x^{2} - 4 x - 5 = 0$
  $(x - 5) (x + 1) = 0$
  $x = 5$ or $x = - 1$
  Therefore $A (- 1; 0)$
  1 mark · Find the x-intercepts
2. ✓
  Calculate gradient of AC $m = \frac{8 - 0}{3 - ( - 1 )} = \frac{8}{4}$
  $m = 2$
  1 mark · Calculate gradient between two points
3. ✓
  Determine equation of line $y - 0 = 2 (x + 1)$
  $y = 2 x + 2$
  1 mark · Find the equation of a line
6. Question 4.3 · March 2015 · Provincial · 3 marks
$P (- 6; 1)$ , $Q (6; - 5)$ and $R (4; 5)$ are the co-ordinates of $Δ PQR$ . C is the midpoint of QR. A, B and C are the intercepts of lines PQ and QR respectively.
$B \hat{Q} C = α$ , $P \hat{A} X = θ_{1}$ and $R \hat{C} X = θ_{2}$ .
4.3 Determine the equation of PQ.(3)
Show the memo (3 ticks)Memo
1. ✓
  Substitution $y = m x + c$
  $1 = - \frac{1}{2} (- 6) + c$
  $1 = 3 + c$
  1 mark · Find the equation of a line
2. ✓
  Solve for c $c = - 2$
  1 mark · Find the equation of a line
3. ✓
  Equation $y = - \frac{1}{2} x - 2$
  1 mark · Find the equation of a line

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