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Analytical Geometry

Grade 11 CAPS: the equation of a line through two points, the equation of a line through one point parallel or perpendicular to a given line, and the inclination of a line.

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Grade 11 CAPS Mathematics

Analytical Geometry

Now that a point can be located and measured on the Cartesian plane, this grade teaches how to write down the actual equation of the line joining two such points — and the two natural spin-offs of that skill: parallel/perpendicular lines through a point, and the inclination (angle) of a line.

Scope check:Official CAPS requirement

Your 4-hour learning route

A short Grade 10 revision, then three genuinely new skills, then combining everything.

25
min
1. Grade 10 revision

Distance, gradient, midpoint, parallel/perpendicular.

45
min
2. Equation of a line

From a gradient and a point, and from two points.

50
min
3. Parallel & perpendicular

Writing a new equation through a given point.

45
min
4. Inclination of a line

The angle a line makes with the x-axis.

35
min
5. Combining the toolkit

Perpendicular bisectors and medians.

Study rule
Grade 11 exams test Grade 10's four formulas cumulatively — you can't find the equation of a line through two points without first calculating its gradient exactly the way you learned in Grade 10.

What CAPS actually asks in Grade 11

Building directly on Grade 10's four formulas.

  • 1

    Solve problems in 2-dimensions. Use a Cartesian co-ordinate system to derive and apply the equation of a line through two given points.

  • 2

    The equation of a line through one point and parallel or perpendicular to a given line.

  • 3

    The inclination of a line.

Carried over from Grade 10
Distance, gradient, midpoint, and the parallel/perpendicular condition are assumed knowledge here — this page briefly revises them, but for the full treatment (with more examples) see the Grade 10 page.

Quick revision: Grade 10's toolkit

Four formulas you'll use constantly this grade — full detail on the Grade 10 page.

Distance
\[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]
Gradient
\[m=\dfrac{y_2-y_1}{x_2-x_1}\]
Midpoint
\[M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\]
Parallel / Perpendicular
\[m_1=m_2 \quad\text{or}\quad m_1\times m_2=-1\]
Not a full re-teach
This is a reminder, not a lesson — if any of these four feel unfamiliar, work through the Grade 10 Summary Notes first.

The equation of a line

Every straight line (that isn't vertical) can be written as \(y=mx+c\).

Gradient-intercept form
\[y=mx+c\]

\(m\) is the gradient; \(c\) is the \(y\)-intercept (where the line crosses the \(y\)-axis, at \(x=0\)).

Finding the equation from a gradient and ONE point
  1. Write \(y=mx+c\), substituting the known gradient for \(m\).
  2. Substitute the point's \(x\) and \(y\) values into the equation.
  3. Solve for \(c\).
  4. Write the final equation with both \(m\) and \(c\) filled in.

Worked example: equation from a gradient and a point

Direct substitution, negative gradient, then a practical follow-up: reading off the intercepts.

Direct substitution Level 1

Find the equation of the line with gradient 2, passing through \((1,3)\).

Show the working
  1. 1\(y=2x+c\)
  2. 2Substitute \((1,3)\): \(3=2(1)+c\Rightarrow c=1\)
  3. 3\(\boxed{y=2x+1}\)
Negative gradient, negative coordinates Level 2

Find the equation of the line with gradient \(-3\), passing through \((-2,5)\).

Show the working
  1. 1\(y=-3x+c\)
  2. 2Substitute \((-2,5)\): \(5=-3(-2)+c=6+c\Rightarrow c=-1\)
  3. 3\(\boxed{y=-3x-1}\)
Application: finding both intercepts Level 2-3

For the line \(y=2x+1\), calculate the coordinates of the \(x\)-intercept and the \(y\)-intercept.

Show the working
  1. 1\(y\)-intercept: set \(x=0\): \(y=2(0)+1=1\), so \((0,1)\) — this is just \(c\) itself.
  2. 2\(x\)-intercept: set \(y=0\): \(0=2x+1\Rightarrow x=-0{,}5\), so \(\boxed{(-0{,}5,0)}\)
Why this works
The \(y\)-intercept is always just \(c\) — no calculation needed once the equation is in \(y=mx+c\) form. The \(x\)-intercept always needs the substitution \(y=0\).
Quick check: equation from gradient + point

Find the equation of the line with gradient 4, passing through \((0,3)\).

On the graph of y = mx + c, what does c represent?

The equation of a line through two points

Two steps: find the gradient first (Grade 10's formula), then find c exactly as before.

Method
  1. Calculate the gradient using the Grade 10 gradient formula.
  2. Substitute EITHER of the two given points into \(y=mx+c\).
  3. Solve for \(c\).

Worked example: equation from two points

Positive gradient, negative gradient, working backward, then a real application.

Direct case Level 1

Find the equation of the line through \(A(1,2)\) and \(B(3,6)\).

Show the working
  1. 1\(m=\dfrac{6-2}{3-1}=2\)
  2. 2Substitute \(A(1,2)\): \(2=2(1)+c\Rightarrow c=0\)
  3. 3\(\boxed{y=2x}\)
Negative gradient Level 2

Find the equation of the line through \(A(-1,4)\) and \(B(3,-4)\).

Show the working
  1. 1\(m=\dfrac{-4-4}{3-(-1)}=\dfrac{-8}{4}=-2\)
  2. 2Substitute \(A(-1,4)\): \(4=-2(-1)+c=2+c\Rightarrow c=2\)
  3. 3\(\boxed{y=-2x+2}\)
Working backward for an unknown coordinate Level 3

The line through \((1,2)\) and \((4,k)\) has equation \(y=3x-1\). Calculate \(k\).

Show the working
  1. 1Since \((4,k)\) lies on the line, substitute \(x=4\) into the given equation: \(k=3(4)-1\)
  2. 2\(\boxed{k=11}\)
Application: is a point on the line? Level 4

\(D(2,1)\) and \(E(6,13)\) define a line. Determine whether \(F(4,7)\) lies on line \(DE\).

Show the working
  1. 1\(m_{DE}=\dfrac{13-1}{6-2}=\dfrac{12}{4}=3\)
  2. 2Substitute \(D(2,1)\): \(1=3(2)+c\Rightarrow c=-5\), so line \(DE\): \(y=3x-5\)
  3. 3Test \(F(4,7)\): \(3(4)-5=7\), which matches \(F\)'s \(y\)-value exactly, so \(\boxed{F\text{ lies on line }DE}\)
Common mistake
You may substitute EITHER given point to solve for \(c\) — both give the same answer. Don't waste time substituting both "to check"; substituting one is a complete, correct method.
Quick check: equation from two points

Find the equation of the line through \((0,0)\) and \((2,8)\).

After calculating the gradient of a line through two given points, what is the next step?

A line through a point, parallel to a given line

Same method as before — the given line only tells you the gradient.

Method
  1. Identify the gradient of the GIVEN line (rearrange to \(y=mx+c\) form first if needed).
  2. A parallel line has the SAME gradient: \(m_{\text{new}}=m_{\text{given}}\).
  3. Substitute the new point into \(y=mx+c\) and solve for the new \(c\).

Worked example: parallel line through a point

A gradient already given, then a general-form equation, then a reverse problem.

Gradient given directly Level 1

Find the equation of the line through \((1,1)\), parallel to \(y=2x+3\).

Show the working
  1. 1Parallel means the same gradient: \(m=2\)
  2. 2Substitute \((1,1)\): \(1=2(1)+c\Rightarrow c=-1\)
  3. 3\(\boxed{y=2x-1}\)
Rearranging a general-form equation first Level 2-3

Find the equation of the line through \((3,-1)\), parallel to \(2x+3y-6=0\).

Show the working
  1. 1Rearrange into \(y=mx+c\) form first: \(3y=-2x+6\Rightarrow y=-\dfrac{2}{3}x+2\), so \(m=-\dfrac{2}{3}\)
  2. 2Substitute \((3,-1)\): \(-1=-\dfrac{2}{3}(3)+c=-2+c\Rightarrow c=1\)
  3. 3\(\boxed{y=-\dfrac{2}{3}x+1}\)
Reverse: solving for an unknown constant Level 4

Line 1: \(y=3x-2\). Line 2: \(kx-2y+5=0\) is PARALLEL to Line 1. Calculate \(k\).

Show the working
  1. 1Rearrange Line 2 into \(y=mx+c\) form: \(-2y=-kx-5\Rightarrow y=\dfrac{k}{2}x+\dfrac{5}{2}\), so its gradient is \(\dfrac{k}{2}\)
  2. 2Parallel means equal gradients: \(\dfrac{k}{2}=3\)
  3. 3\(\boxed{k=6}\)

A line through a point, perpendicular to a given line

Same method again, using the negative-reciprocal gradient instead.

Method
  1. Identify the gradient of the GIVEN line.
  2. A perpendicular line's gradient is the negative reciprocal: \(m_{\text{new}}=-\dfrac{1}{m_{\text{given}}}\).
  3. Substitute the new point into \(y=mx+c\) and solve for the new \(c\).

Worked example: perpendicular line through a point

A gradient already given, then a general-form equation, then a reverse problem.

Gradient given directly Level 1

Find the equation of the line through \((4,1)\), perpendicular to \(y=2x+3\).

Show the working
  1. 1Perpendicular gradient: \(m=-\dfrac{1}{2}\)
  2. 2Substitute \((4,1)\): \(1=-\dfrac{1}{2}(4)+c=-2+c\Rightarrow c=3\)
  3. 3\(\boxed{y=-\dfrac{1}{2}x+3}\)
Rearranging a general-form equation first Level 2-3

Find the equation of the line through \((2,5)\), perpendicular to \(2x+3y-6=0\).

Show the working
  1. 1Rearrange first: \(y=-\dfrac{2}{3}x+2\), so \(m=-\dfrac{2}{3}\)
  2. 2Perpendicular gradient: \(m_{\perp}=-\dfrac{1}{-2/3}=\dfrac{3}{2}\)
  3. 3Substitute \((2,5)\): \(5=\dfrac{3}{2}(2)+c=3+c\Rightarrow c=2\)
  4. 4\(\boxed{y=\dfrac{3}{2}x+2}\)
Reverse: solving for an unknown constant Level 4

Line 1: \(y=3x-2\). Line 2: \(kx+y-4=0\) is PERPENDICULAR to Line 1. Calculate \(k\).

Show the working
  1. 1Rearrange Line 2: \(y=-kx+4\), so its gradient is \(-k\)
  2. 2Perpendicular condition: \(3\times(-k)=-1\)
  3. 3\(\boxed{k=\dfrac{1}{3}}\)
Quick check: parallel & perpendicular lines

Find the equation of the line through (2,7), parallel to y = 4x - 1.

A line has gradient 5. A second line, kx + y = 2, is perpendicular to it. What is k?

The inclination of a line

The angle a line makes with the positive x-axis.

Inclination and gradient
\[m=\tan\theta\]

\(\theta\) (theta) is measured anticlockwise from the positive \(x\)-axis, and always lies between \(0°\) and \(180°\). A positive gradient gives an acute angle (\(0°<\theta<90°\)); a negative gradient gives an obtuse angle (\(90°<\theta<180°\)).

Worked example: inclination of a line

Gradient to angle (both signs), angle to gradient, then a genuine application.

Positive gradient Level 1

Calculate the inclination of a line with gradient 1.

Show the working
\(\tan\theta=1\Rightarrow\boxed{\theta=45°}\)
Negative gradient Level 2

Calculate the inclination of a line with gradient \(-1\).

Show the working
  1. 1\(\tan\theta=-1\). Your calculator gives \(-45°\), but inclination must be between \(0°\) and \(180°\)
  2. 2Add \(180°\) to bring it into range: \(-45°+180°=\boxed{135°}\)
Working backward: angle to gradient Level 2

A line has an inclination of \(60°\). Calculate its gradient.

Show the working
\(m=\tan60°=\boxed{\sqrt{3}\approx1{,}73}\)
Application: are two lines perpendicular? Level 4

Line 1 has an inclination of \(30°\). Line 2 has an inclination of \(120°\). Are the two lines perpendicular?

Show the working
  1. 1\(m_1=\tan30°=\dfrac{1}{\sqrt3}\approx0{,}577\)
  2. 2\(m_2=\tan120°=-\sqrt3\approx-1{,}732\)
  3. 3\(m_1\times m_2=\dfrac{1}{\sqrt3}\times(-\sqrt3)=\boxed{-1}\), so yes — the lines are perpendicular
Common mistake
A calculator's \(\tan^{-1}\) button always returns a value between \(-90°\) and \(90°\). Whenever the gradient is negative, that raw calculator answer is NEGATIVE — you must add \(180°\) to place it correctly in the \(0°\)-\(180°\) inclination range.
Quick check: inclination

Calculate the inclination of a line with gradient \(\sqrt{3}\).

A calculator gives tan^-1(-2) = -63,4°. What is the correct inclination?

Combining the toolkit

Real exam questions often ask for a special line related to a triangle or segment.

Two classic combinations

Perpendicular bisector of segment \(AB\): find the midpoint of \(AB\) (Grade 10), then find the equation of the line through that midpoint with the perpendicular gradient to \(AB\).

Median of a triangle (from a vertex to the midpoint of the opposite side): find the midpoint of the opposite side, then find the equation of the line through the vertex and that midpoint.

Worked example: the perpendicular bisector

Midpoint formula, perpendicular gradient, then the equation of a line — three Grade 10/11 skills in one problem.

Find the perpendicular bisector of AB Level 4

\(A(-2,1)\) and \(B(4,7)\). Find the equation of the perpendicular bisector of \(AB\).

Show the complete solution
  1. 1Midpoint of \(AB\): \(M=\left(\dfrac{-2+4}{2},\dfrac{1+7}{2}\right)=(1,4)\)
  2. 2Gradient of \(AB\): \(m_{AB}=\dfrac{7-1}{4-(-2)}=\dfrac{6}{6}=1\)
  3. 3Perpendicular gradient: \(m_{\perp}=-\dfrac{1}{1}=-1\)
  4. 4Substitute \(M(1,4)\) into \(y=-x+c\): \(4=-1+c\Rightarrow c=5\)
  5. 5\(\boxed{y=-x+5}\)

Worked example: the median of a triangle

Combining the midpoint formula with the equation of a line through two points.

Find the median from a vertex Level 4

\(\triangle ABC\) has vertices \(A(2,9)\), \(B(-4,1)\) and \(C(6,-3)\). Find the equation of the median from \(A\) to the midpoint of \(BC\).

Show the working
  1. 1Midpoint of \(BC\): \(D=\left(\dfrac{-4+6}{2},\dfrac{1+(-3)}{2}\right)=(1,-1)\)
  2. 2Gradient of \(AD\): \(m_{AD}=\dfrac{-1-9}{1-2}=\dfrac{-10}{-1}=10\)
  3. 3Substitute \(A(2,9)\): \(9=10(2)+c=20+c\Rightarrow c=-11\)
  4. 4\(\boxed{y=10x-11}\)
Read the question carefully
A median always goes from a vertex to the MIDPOINT of the OPPOSITE side — the median "from A" uses the midpoint of BC, never AB or AC.

Mistake clinic: repair the exact error

Small slips that break an otherwise correct answer.

Equation from two points

Substitute AFTER finding m

Find the gradient first, then substitute a point — never substitute a point into an equation before you know \(m\).

General form

Rearrange fully first

\(Ax+By+C=0\) does not show the gradient directly — always rearrange to \(y=mx+c\) before reading off \(m\).

Perpendicular

Flip AND change sign

The negative reciprocal needs both steps — flipping only, or negating only, is not enough.

Inclination

Fix a negative calculator answer

A negative \(\tan^{-1}\) result needs \(180°\) added to land in the valid \(0°\)-\(180°\) range.

Median

Opposite side, not adjacent

The median from a vertex always uses the midpoint of the side that vertex does NOT touch.

Your Paper 2 checklist

Use this before submitting an exam answer.

Avoid this

Forgetting Grade 10's gradient formula is the first step to finding an equation from two points.

Reading \(m\) directly off a general-form equation without rearranging first.

Leaving a negative inclination answer instead of adding \(180°\).

Do this

Always write the final answer in full \(y=mx+c\) form unless asked otherwise.

Sketch the line roughly — a sketch catches an obviously wrong gradient or intercept immediately.

For "median" or "perpendicular bisector" questions, identify the midpoint FIRST before anything else.

Find the equation of a line given its gradient and one point.
Find the equation of a line through two given points, including working backward for an unknown coordinate.
Find the equation of a line through a point, parallel to a given line, including from general form.
Find the equation of a line through a point, perpendicular to a given line, including solving for an unknown constant.
Calculate the inclination of a line from its gradient, and vice versa.
Combine these skills to find a perpendicular bisector or a median.
More explanation and exercises:Siyavula Grade 11 Analytical Geometry
Summary complete

You now have the toolkit.

Use the Mastery Bank to build fluency. Then take the test without notes and use the result to choose the exact slide to revisit.

Next step 1

Mastery Bank

17 questions by skill, from a direct equation calculation through to a perpendicular bisector. Answers reveal only after your attempt.

Next step 2

Quick Test

A short exam-style self-check. Mark it, then revisit the exact slide that matches a missed skill.

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Learn It in Short Videos

Four free, independent videos — not made by Equation Station SA.

Equation From Gradient + Point

Writing an equation from a slope and one point.

Khan Academy · Introduction to point-slope form

Equation From Two Points

Finding the gradient first, then writing the equation.

Khan Academy · Point-slope and slope-intercept form from two points

Parallel & Perpendicular Through a Point

Writing a new equation perpendicular to a given line — the same method works for parallel lines using equal gradients instead.

Khan Academy · Writing equations of perpendicular lines

Inclination of a Line

A CAPS-specific explanation of the angle-gradient relationship.

Kevinmathscience · Grade 11 Analytical geometry: angle of inclination

Practise in the right order

The core teaching is above. These are the next steps, not a replacement for it.

01
Built-in practice
Analytical Geometry Mastery Bank

17 questions by skill, with concise reveal answers and methods.

Start after the slides
Open Mastery Bank
02
Built-in check
Test Your Knowledge

Use the short exam-style self-check when you want a fast confidence check.

Then target one weak skill
Take the Quick Test
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Siyavula: Grade 11 Analytical Geometry

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Related Topics

This topic continues from Grade 10 and into Grade 12.

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Analytical Geometry (Grade 10)

Distance, gradient, midpoint, parallel/perpendicular

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Analytical Geometry (Grade 12)

Circles and tangents

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Functions & Graphs (Grade 11)

Parameters and average gradient

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Student Questions

Frequently Asked Questions

Short answers for the checks learners make while preparing for the Grade 11 CAPS exam.

What does CAPS require for Grade 11 Analytical Geometry?

Solve problems in 2 dimensions. Use a Cartesian co-ordinate system to derive and apply the equation of a line through two given points; the equation of a line through one point and parallel or perpendicular to a given line; and the inclination of a line.

What is the inclination of a line?

The inclination of a line is the angle it makes with the positive direction of the x-axis, measured anticlockwise, between 0 degrees and 180 degrees. It relates to the gradient by m = tan(theta).

Do I need Grade 10's formulas for Grade 11 Analytical Geometry?

Yes. Grade 11 exams test the distance, gradient, midpoint, and parallel/perpendicular condition cumulatively alongside the new Grade 11 content, since finding the equation of a line through two points starts by calculating the gradient between them.

Where should I practise next?

Finish the interactive slides, open the Mastery Bank, then take the short self-test without notes.