Coordinate Geometry.

Grade 11 Mathematics: Section 1 - Coordinate Geometry: Distance, Midpoint, Slope and Line Equations

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Lesson Objectives

  • Understand the Cartesian plane as a framework for locating and relating points
  • Apply the distance formula to find the length of a line segment joining two points
  • Apply the midpoint formula to find the point exactly halfway between two given points
  • Calculate the slope (gradient) of a line joining two points and interpret its sign
  • Determine whether two lines are parallel or perpendicular using their slopes
  • Find the equation of a straight line given sufficient information about it
  • Apply these coordinate geometry tools together to solve multi-step problems

Introduction to Coordinate Geometry

Coordinate geometry (also called analytic geometry) connects algebra and geometry by describing geometric shapes and relationships using numbers and equations on the Cartesian plane. Instead of relying on rulers and protractors, we can calculate exact lengths, midpoints, and angles of inclination using formulas. These tools are used extensively in fields such as engineering (calculating structural distances and slopes), navigation (plotting the shortest route between two locations), computer graphics (positioning and connecting points on a screen), and architecture (checking that walls are parallel or perpendicular). This lesson brings together five closely related coordinate geometry tools — distance, midpoint, slope, parallelism/perpendicularity, and the equation of a line — that form the foundation for almost every other topic in coordinate geometry.

Key Formulas for This Lesson
Distance: $$ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} $$ Midpoint: $$ M = \left( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right) $$ Slope: $$ m = \dfrac{y_2 - y_1}{x_2 - x_1} $$ Parallel lines: $$ m_1 = m_2 $$ Perpendicular lines: $$ m_1 \times m_2 = -1 $$ Equation of a line (point-slope form): $$ y - y_1 = m(x - x_1) $$
Key Definitions:
• Line segment: a straight path between two fixed points, including the points themselves.
• Slope (gradient): a measure of the steepness and direction of a line, calculated as the ratio of vertical change to horizontal change.
• Parallel lines: two straight lines that never meet, because they have exactly the same slope.
• Perpendicular lines: two straight lines that intersect at a right angle ($90^\circ$), where the product of their slopes is $-1$.
• Collinear points: points that lie on the same straight line, which can be verified by checking that the slope between every pair of points is the same.

Distance Between Two Points

The distance between two points on the Cartesian plane is the length of the straight line segment joining them. This formula comes directly from the Pythagorean theorem: the horizontal and vertical separations between the points form the two shorter sides of a right triangle, and the segment joining the points is the hypotenuse. $$ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} $$

Step-by-Step Method:
1. Label the two given points as $(x_1, y_1)$ and $(x_2, y_2)$.
2. Find the horizontal difference $(x_2 - x_1)$ and the vertical difference $(y_2 - y_1)$.
3. Square each difference, then add the two squared values together.
4. Take the square root of the sum to obtain the distance.
Example 1: Basic Distance Calculation
Problem: Find the distance between $A(1, 2)$ and $B(4, 6)$.

Formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Solution:
Step 1: $(x_1,y_1)=(1,2)$, $(x_2,y_2)=(4,6)$.

Step 2: $x_2-x_1 = 4-1 = 3$, $y_2-y_1 = 6-2 = 4$.

Step 3: $d = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25}$.

Answer: $d = 5$ units
Example 2: Distance with Negative Coordinates
Problem: Find the distance between $P(-2, -3)$ and $Q(4, 5)$, leaving your answer in surd form if it is not exact.

Formula: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Solution:
Step 1: $(x_1,y_1)=(-2,-3)$, $(x_2,y_2)=(4,5)$.

Step 2: $x_2-x_1 = 4-(-2) = 6$, $y_2-y_1 = 5-(-3) = 8$.

Step 3: $d = \sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100}$.

Answer: $d = 10$ units
Watch Out!
The order in which you subtract the coordinates does not matter for distance, since the difference is squared in both cases. However, be consistent within a single calculation — do not mix $(x_1 - x_2)$ in one term with $(x_2 - x_1)$ in the other, as this can cause sign errors when the differences are not squared correctly.

Practice for Distance Between Two Points

  1. Find the distance between $(0, 0)$ and $(3, 4)$.
  2. Find the distance between $(2, 5)$ and $(2, 9)$, and explain why this can also be found without the full formula.
  3. Find the distance between $(-1, 2)$ and $(5, -6)$.
  4. Find the distance between $(3, 7)$ and $(8, 3)$, leaving your answer in surd form.
  5. A triangle has vertices $(0,0)$, $(6,0)$, and $(0,8)$. Find the length of each side.

Midpoint of a Line Segment

The midpoint of a line segment is the point that lies exactly halfway between its two endpoints. It is found by averaging the x-coordinates and averaging the y-coordinates of the two endpoints separately. $$ M = \left( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right) $$

Step-by-Step Method:
1. Label the two given points as $(x_1, y_1)$ and $(x_2, y_2)$.
2. Add the x-coordinates and divide by 2 to obtain the midpoint's x-coordinate.
3. Add the y-coordinates and divide by 2 to obtain the midpoint's y-coordinate.
4. Write the final answer as an ordered pair $(x, y)$, simplifying any fractions.
Example 3: Midpoint with Mixed Signs
Problem: Find the midpoint of the segment joining $A(-4, 7)$ and $B(6, -3)$.

Formula: $M = \left( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right)$

Solution:
Step 1: $(x_1,y_1)=(-4,7)$, $(x_2,y_2)=(6,-3)$.

Step 2: $x = \dfrac{-4+6}{2} = 1$.

Step 3: $y = \dfrac{7+(-3)}{2} = 2$.

Answer: $M(1, 2)$
Example 4: Finding a Missing Endpoint from a Midpoint
Problem: The midpoint of $EF$ is $M(3, 1)$. If $E(-1, 4)$, find the coordinates of $F$.

Formula: $x_2 = 2x_M - x_1$, $y_2 = 2y_M - y_1$

Solution:
Step 1: $x_M = \dfrac{x_1+x_2}{2} \Rightarrow 3 = \dfrac{-1+x_2}{2} \Rightarrow 6 = -1+x_2 \Rightarrow x_2 = 7$.

Step 2: $y_M = \dfrac{y_1+y_2}{2} \Rightarrow 1 = \dfrac{4+y_2}{2} \Rightarrow 2 = 4+y_2 \Rightarrow y_2 = -2$.

Answer: $F(7, -2)$
Watch Out!
Do not round fractional midpoint coordinates unless the question asks for a decimal or approximate answer. An exact fraction such as $\dfrac{11}{2}$ is usually preferred over $5.5$ in SSCE marking schemes unless stated otherwise.

Practice for Midpoint of a Line Segment

  1. Find the midpoint of the segment joining $(2, 3)$ and $(10, 9)$.
  2. Find the midpoint of the segment joining $(-6, -2)$ and $(6, 2)$.
  3. The midpoint of $GH$ is $(4, -2)$. If $G(-2, 3)$, find $H$.
  4. Find the midpoint of the segment joining $(3, 4)$ and $(4, 7)$, leaving your answer in fractional form.
  5. A triangle has vertices $(0,0)$, $(6,0)$, and $(0,8)$. Find the midpoint of the side joining $(6,0)$ and $(0,8)$.

Slope (Gradient) of a Line

The slope (or gradient) of a line measures how steep it is and in which direction it inclines. It is defined as the ratio of the vertical change (rise) to the horizontal change (run) between any two points on the line. The sign and size of the slope tell us a great deal about the line's direction without needing to see a graph. $$ m = \dfrac{y_2 - y_1}{x_2 - x_1} $$

Step-by-Step Method:
1. Label the two given points as $(x_1, y_1)$ and $(x_2, y_2)$.
2. Find the vertical change $(y_2 - y_1)$ and the horizontal change $(x_2 - x_1)$.
3. Divide the vertical change by the horizontal change, keeping the coordinates in matching order.
4. Simplify the resulting fraction, and interpret its sign (rising, falling, flat, or vertical).
Example 5: Calculating a Positive Slope
Problem: Find the slope of the line joining $A(1, 2)$ and $B(5, 10)$.

Formula: $m = \dfrac{y_2-y_1}{x_2-x_1}$

Solution:
Step 1: $(x_1,y_1)=(1,2)$, $(x_2,y_2)=(5,10)$.

Step 2: $y_2-y_1 = 10-2 = 8$, $x_2-x_1 = 5-1 = 4$.

Step 3: $m = \dfrac{8}{4} = 2$.

Answer: $m = 2$ (the line rises 2 units for every 1 unit moved to the right)
Example 6: Calculating a Negative Slope
Problem: Find the slope of the line joining $C(-3, 6)$ and $D(2, -4)$.

Formula: $m = \dfrac{y_2-y_1}{x_2-x_1}$

Solution:
Step 1: $(x_1,y_1)=(-3,6)$, $(x_2,y_2)=(2,-4)$.

Step 2: $y_2-y_1 = -4-6 = -10$, $x_2-x_1 = 2-(-3) = 5$.

Step 3: $m = \dfrac{-10}{5} = -2$.

Answer: $m = -2$ (the line falls as it moves to the right)

Classifying Slopes

Slope ValueTypeLine DirectionExample Points
$m > 0$PositiveRises from left to right$(1,2)$ and $(5,10)$
$m < 0$NegativeFalls from left to right$(-3,6)$ and $(2,-4)$
$m = 0$ZeroHorizontal (flat) line$(2,4)$ and $(7,4)$
$m$ undefinedUndefinedVertical line$(3,1)$ and $(3,8)$

Scroll sideways on smaller screens to view the full table.

Watch Out!
A slope is undefined (not zero) when the line is vertical, because this makes the denominator $(x_2-x_1)$ equal to zero, and division by zero is not possible. Do not confuse "zero slope" (a flat, horizontal line) with "undefined slope" (a vertical line) — they describe opposite situations.

Practice for Slope (Gradient) of a Line

  1. Find the slope of the line joining $(0, 0)$ and $(4, 8)$.
  2. Find the slope of the line joining $(2, 5)$ and $(2, -3)$, and state what this tells you about the line.
  3. Find the slope of the line joining $(-5, 3)$ and $(1, 3)$, and state what this tells you about the line.
  4. Find the slope of the line joining $(-2, -7)$ and $(3, 8)$.
  5. Three points $(1,2)$, $(3,6)$, and $(5,10)$ are claimed to be collinear. Use slope to verify whether this is true.

Parallel and Perpendicular Lines

Once we can calculate slope, we can use it to compare two lines. Two lines are parallel if they have exactly the same slope, since they rise and fall at the same rate and therefore never meet. Two lines are perpendicular if they meet at a right angle, which happens precisely when the product of their slopes equals $-1$ — in other words, one slope is the negative reciprocal of the other. $$ \text{Parallel: } m_1 = m_2 \qquad \text{Perpendicular: } m_1 \times m_2 = -1 $$

Step-by-Step Method:
1. Calculate the slope of each line separately using $m = \dfrac{y_2-y_1}{x_2-x_1}$.
2. To test for parallel lines, check whether $m_1 = m_2$.
3. To test for perpendicular lines, multiply the two slopes together and check whether $m_1 \times m_2 = -1$.
4. State your conclusion clearly, referencing the calculated slope values.
Example 7: Testing for Parallel Lines
Problem: Line $AB$ passes through $A(1,1)$ and $B(4,7)$. Line $CD$ passes through $C(2,-1)$ and $D(5,5)$. Determine whether $AB$ is parallel to $CD$.

Formula: $m = \dfrac{y_2-y_1}{x_2-x_1}$; Parallel condition: $m_1 = m_2$

Solution:
Step 1: Slope of $AB$: $m_1 = \dfrac{7-1}{4-1} = \dfrac{6}{3} = 2$.

Step 2: Slope of $CD$: $m_2 = \dfrac{5-(-1)}{5-2} = \dfrac{6}{3} = 2$.

Step 3: Since $m_1 = m_2 = 2$, the two lines have equal slopes.

Answer: $AB$ is parallel to $CD$
Example 8: Testing for Perpendicular Lines
Problem: Line $PQ$ passes through $P(0,0)$ and $Q(2,4)$. Line $RS$ passes through $R(1,5)$ and $S(5,3)$. Determine whether $PQ$ is perpendicular to $RS$.

Formula: $m = \dfrac{y_2-y_1}{x_2-x_1}$; Perpendicular condition: $m_1 \times m_2 = -1$

Solution:
Step 1: Slope of $PQ$: $m_1 = \dfrac{4-0}{2-0} = 2$.

Step 2: Slope of $RS$: $m_2 = \dfrac{3-5}{5-1} = \dfrac{-2}{4} = -\dfrac{1}{2}$.

Step 3: $m_1 \times m_2 = 2 \times \left(-\dfrac{1}{2}\right) = -1$.

Answer: $PQ$ is perpendicular to $RS$

Parallel vs Perpendicular: Quick Comparison

RelationshipSlope Condition$m_1$$m_2$Check
Parallel$m_1 = m_2$$2$$2$Equal, so parallel
Perpendicular$m_1 \times m_2 = -1$$2$$-\dfrac{1}{2}$$2 \times -\dfrac{1}{2}=-1$, so perpendicular
Neither$m_1 \neq m_2$ and $m_1 \times m_2 \neq -1$$3$$1$Not equal; product is $3$, not $-1$

Scroll sideways on smaller screens to view the full table.

Watch Out!
To find the slope of a line perpendicular to a line of slope $m$, take the negative reciprocal: flip the fraction and change its sign. A common error is only flipping the fraction (forgetting to change the sign) or only changing the sign (forgetting to flip it).

Practice for Parallel and Perpendicular Lines

  1. Show that the line joining $(0,0)$ and $(3,6)$ is parallel to the line joining $(1,1)$ and $(4,7)$.
  2. Show that the line joining $(2,3)$ and $(6,3)$ is perpendicular to the line joining $(4,-1)$ and $(4,5)$.
  3. Find the slope of a line perpendicular to a line with slope $\dfrac{3}{4}$.
  4. Determine whether the line joining $(-1,2)$ and $(3,10)$ is parallel, perpendicular, or neither to the line joining $(0,5)$ and $(4,7)$.
  5. A quadrilateral has one pair of sides with slopes $\dfrac{2}{3}$ and $\dfrac{2}{3}$, and another pair with slopes $-\dfrac{3}{2}$ and $-\dfrac{3}{2}$. What can you conclude about the shape?

Equation of a Straight Line

Once we know the slope of a line and at least one point it passes through, we can write down its equation. The point-slope form is the most direct way to do this, and it can always be rearranged into the more familiar slope-intercept form $y = mx + c$. $$ y - y_1 = m(x - x_1) $$

Step-by-Step Method:
1. Find the slope $m$ of the line (calculate it if not already given).
2. Choose one known point $(x_1, y_1)$ on the line.
3. Substitute $m$, $x_1$, and $y_1$ into $y - y_1 = m(x-x_1)$.
4. Expand and simplify into the form $y = mx + c$, or another form requested by the question.
Example 9: Equation from Two Points
Problem: Find the equation of the line passing through $A(1, 2)$ and $B(4, 11)$, giving your answer in the form $y = mx + c$.

Formula: $m = \dfrac{y_2-y_1}{x_2-x_1}$; $y-y_1=m(x-x_1)$

Solution:
Step 1: $m = \dfrac{11-2}{4-1} = \dfrac{9}{3} = 3$.

Step 2: Use point $A(1,2)$: $y - 2 = 3(x-1)$.

Step 3: Expand: $y - 2 = 3x - 3$.

Step 4: Simplify: $y = 3x - 1$.

Answer: $y = 3x - 1$
Example 10: Equation of a Line Perpendicular to a Given Line
Problem: Find the equation of the line that passes through $(2, 5)$ and is perpendicular to the line $y = 2x + 1$.

Formula: Perpendicular slope $= -\dfrac{1}{m}$; $y-y_1=m(x-x_1)$

Solution:
Step 1: The given line $y=2x+1$ has slope $m=2$.

Step 2: The perpendicular slope is the negative reciprocal: $-\dfrac{1}{2}$.

Step 3: Substitute point $(2,5)$ and slope $-\dfrac{1}{2}$: $y - 5 = -\dfrac{1}{2}(x-2)$.

Step 4: Expand and simplify: $y - 5 = -\dfrac{1}{2}x + 1 \Rightarrow y = -\dfrac{1}{2}x + 6$.

Answer: $y = -\dfrac{1}{2}x + 6$
Watch Out!
When a question gives the equation of a line in a form other than $y = mx + c$ (for example $2x + 4y = 8$), rearrange it into $y = mx + c$ first before reading off the slope — the coefficient of $x$ before rearranging is not the slope.

Practice for Equation of a Straight Line

  1. Find the equation of the line passing through $(0, 3)$ with slope $4$.
  2. Find the equation of the line passing through $(2, 1)$ and $(6, 9)$.
  3. Find the equation of the line through $(-1, 4)$ that is parallel to $y = 3x - 2$.
  4. Find the equation of the line through $(3, -2)$ that is perpendicular to $y = -\dfrac{1}{3}x + 5$.
  5. Rearrange $3x + 2y = 12$ into the form $y = mx + c$, and state its slope.

Cumulative Practice Exercises

  1. Find the distance between $(3, -2)$ and $(-1, 1)$.
  2. Find the midpoint of the segment joining $(-5, 8)$ and $(7, -2)$.
  3. Find the slope of the line joining $(4, -1)$ and $(-2, 5)$.
  4. The midpoint of $XY$ is $(2, 2)$. If $X(-4, 6)$, find $Y$.
  5. Determine whether the line joining $(0,0)$ and $(4,2)$ is parallel to the line joining $(1,3)$ and $(5,5)$.
  6. Determine whether the line joining $(1,1)$ and $(3,5)$ is perpendicular to the line joining $(2,4)$ and $(4,3)$.
  7. Find the equation of the line passing through $(1, -3)$ with slope $-2$.
  8. Find the distance between $(0, 5)$ and $(12, 0)$.
  9. A triangle has vertices $A(0,0)$, $B(4,0)$, $C(0,3)$. Find the length of each side and classify the triangle by its side lengths.
  10. Find the equation of the line through $(2, 3)$ that is perpendicular to the line $2x + y = 5$.
  11. Show that the points $(1,2)$, $(3,6)$, and $(6,12)$ are collinear using slope.
  12. Find the midpoint of the diagonal joining $(-3, -1)$ and $(9, 7)$ in quadrilateral $PQRS$, and state what this represents if $PQRS$ is known to be a parallelogram.
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Solutions to Cumulative Exercises

  1. Step 1: $x_2-x_1 = -1-3 = -4$, $y_2-y_1 = 1-(-2) = 3$.
    Step 2: $d = \sqrt{(-4)^2+3^2} = \sqrt{16+9} = \sqrt{25}$.
    Answer: $d = 5$ units
  2. Step 1: $x = \dfrac{-5+7}{2} = 1$.
    Step 2: $y = \dfrac{8+(-2)}{2} = 3$.
    Answer: $(1, 3)$
  3. Step 1: $y_2-y_1 = 5-(-1) = 6$, $x_2-x_1 = -2-4 = -6$.
    Step 2: $m = \dfrac{6}{-6} = -1$.
    Answer: $m = -1$
  4. Step 1: $x_M = \dfrac{x_1+x_2}{2} \Rightarrow 2 = \dfrac{-4+x_2}{2} \Rightarrow x_2 = 8$.
    Step 2: $y_M = \dfrac{y_1+y_2}{2} \Rightarrow 2 = \dfrac{6+y_2}{2} \Rightarrow y_2 = -2$.
    Answer: $Y(8, -2)$
  5. Step 1: Slope of first line: $m_1 = \dfrac{2-0}{4-0} = \dfrac{1}{2}$.
    Step 2: Slope of second line: $m_2 = \dfrac{5-3}{5-1} = \dfrac{1}{2}$.
    Answer: $m_1 = m_2 = \dfrac{1}{2}$, so the lines are parallel
  6. Step 1: Slope of first line: $m_1 = \dfrac{5-1}{3-1} = 2$.
    Step 2: Slope of second line: $m_2 = \dfrac{3-4}{4-2} = -\dfrac{1}{2}$. Product: $2 \times -\dfrac{1}{2} = -1$.
    Answer: The lines are perpendicular
  7. Step 1: Substitute into $y-y_1=m(x-x_1)$: $y-(-3) = -2(x-1)$.
    Step 2: Expand and simplify: $y+3 = -2x+2 \Rightarrow y = -2x -1$.
    Answer: $y = -2x - 1$
  8. Step 1: $x_2-x_1 = 12-0 = 12$, $y_2-y_1 = 0-5 = -5$.
    Step 2: $d = \sqrt{12^2+(-5)^2} = \sqrt{144+25} = \sqrt{169}$.
    Answer: $d = 13$ units
  9. Step 1: $AB = \sqrt{(4-0)^2+(0-0)^2} = 4$. $AC = \sqrt{(0-0)^2+(3-0)^2} = 3$.
    Step 2: $BC = \sqrt{(0-4)^2+(3-0)^2} = \sqrt{16+9} = 5$.
    Answer: Sides are $3$, $4$, and $5$ units — a scalene, right-angled triangle (since $3^2+4^2=5^2$)
  10. Step 1: Rearranging $2x+y=5$ gives $y = -2x+5$, so its slope is $-2$; the perpendicular slope is $\dfrac{1}{2}$.
    Step 2: Substitute point $(2,3)$: $y-3 = \dfrac{1}{2}(x-2) \Rightarrow y = \dfrac{1}{2}x + 2$.
    Answer: $y = \dfrac{1}{2}x + 2$
  11. Step 1: Slope from $(1,2)$ to $(3,6)$: $m_1 = \dfrac{6-2}{3-1} = 2$.
    Step 2: Slope from $(3,6)$ to $(6,12)$: $m_2 = \dfrac{12-6}{6-3} = 2$. Since $m_1 = m_2$, the points lie on the same line.
    Answer: The points are collinear
  12. Step 1: $x = \dfrac{-3+9}{2} = 3$.
    Step 2: $y = \dfrac{-1+7}{2} = 3$.
    Answer: $(3, 3)$ — this is the point where both diagonals of parallelogram $PQRS$ meet, since diagonals of a parallelogram bisect each other

Conclusion & Summary

This lesson brought together five essential tools of coordinate geometry: distance, midpoint, slope, the parallel/perpendicular slope conditions, and the equation of a straight line. These tools rarely appear in isolation on SSCE papers — a single question often requires you to calculate a slope, then use it to find an equation, or find a midpoint and then test for perpendicularity. Being fluent in all five formulas, and recognising which one a question is really asking for, will greatly strengthen your coordinate geometry performance.

Key Takeaways:
1. Distance Formula: $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
2. Midpoint Formula: $M = \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$
3. Slope Formula: $m = \dfrac{y_2-y_1}{x_2-x_1}$
4. Parallel/Perpendicular Conditions: $m_1=m_2$ (parallel); $m_1 \times m_2 = -1$ (perpendicular)
5. Equation of a Line: $y - y_1 = m(x-x_1)$

These five formulas work together as a single toolkit — practice switching between them until using the right one becomes automatic.

Video Resource

Watch this video for more examples of slope and coordinate geometry problems.

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