Coordinate Geometry

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Coordinates

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Key Idea — Coordinates

A point in the plane is written as \( (x, y) \).

Coordinate Grid Quiz — Name the Point

Select the correct ordered pair for the point shown.

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Midpoint

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Key Formula — Midpoint

The midpoint of two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is:

\( M\left(\dfrac{x_1 + x_2}{2},\; \dfrac{y_1 + y_2}{2}\right) \)

Midpoint Coordinates

Explore how the midpoint formula works by adjusting the endpoints.
Use this activity to understand how coordinates change and how to find missing endpoints.

Midpoint of a Line Segment

Geometry

Find \(C\), the midpoint of the line segment \([AB]\).

Coordinate Grid Quiz — Midpoint of AB

Two points \(A\) and \(B\) are shown. Select the correct midpoint \(M\).

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A B

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Length and Midpoint of a Line Segment

If \( A(-1,3) \) and \( B(5,7) \) are two points in the plane, find:
(i) \( |AB| \)  (ii) the midpoint of \( [AB] \).

Show worked solution

(i) Find the length \( |AB| \)
Using the distance formula:
\( |AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

\( = \sqrt{(5 - (-1))^2 + (7 - 3)^2} =

\sqrt{6^2 + 4^2} = \sqrt{52} = 2\sqrt{13} \)

(ii) Find the midpoint \( M \)
Midpoint formula:
\( M = \left(\dfrac{x_1 + x_2}{2},\, \dfrac{y_1 + y_2}{2}\right) \)

\( = \left(\dfrac{-1 + 5}{2},\, \dfrac{3 + 7}{2}\right) = (2,5) \)

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Distance

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Key Formula — Distance

The distance between two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is:

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

The Distance Formula from Pythagoras’ Theorem

\[ d = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}. \]

Use the GeoGebra applet to explore how the distance between two points is derived from Pythagoras’ theorem.

Using the distance formula:

For each question, you are given two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\). Fill in the correct values of \(x_1\), \(x_2\), \(y_1\) and \(y_2\) in:

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

d = √( ( − )² + ( − )² )

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Distance Between Two Points — Quiz

Distance formula

For any two points \((x_{1},y_{1})\) and \((x_{2},y_{2})\), \[ d = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}. \]

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Q1. The points are \(A(1,2)\) and \(B(6,5)\). Hint

Use the hint to see the right-angled triangle with \(AB\) as the hypotenuse.

What is the distance \(AB\) between \(A(1,2)\) and \(B(6,5)\)?

A. \(\sqrt{34}\) B. \(\sqrt{25}\) C. \(\sqrt{52}\) D. \(7\)

Slope

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Key Formula — Slope of a Line

The slope of the line through \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is:

\( m=\dfrac{y_2 - y_1}{x_2 - x_1} \)

Understanding Slope

Coordinate Geometry

Move the points, change the line, and explore how the slope ratio behaves.

Using the slope formula:

For each question, you are given two points \(A(x_1,y_1)\) and \(B(x_2,y_2)\). Fill in the correct values of \(x_1\), \(x_2\), \(y_1\) and \(y_2\) in:

\(m = \dfrac{y_2 - y_1}{x_2 - x_1}\).

m =

( − )

( − )

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Coordinate Grid Quiz — Slope (Gradient) of AB

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A B

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Parallel and perpendicular slopes — Quiz

Suppose the slope is \( m=\dfrac{a}{b} \).

Parallel slope: \( m_{\parallel}=\dfrac{a}{b} \)

Perpendicular slope: \( m_{\perp}=-\dfrac{b}{a} \)

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Finding the Slope from a Line Equation

Find the slope (gradient) of the line
\(3x-2y-9=0\).

Show worked solution

Step 1: Rearrange to slope–intercept form.
Start with \(3x-2y-9=0\).

Move the \(x\) and constant terms to the right-hand side:
\(-2y=-3x+9\).

Divide both sides by \(-2\):
\(y=\dfrac{-3}{-2}x+\dfrac{9}{-2} =\dfrac{3}{2}x-\dfrac{9}{2}.\)

Step 2: Read off the slope.
In the form \(y=mx+c\), the coefficient of \(x\) is the slope \(m\).

Here \(m=\dfrac{3}{2}\).

So the slope (gradient) of the line is
\(\boxed{\dfrac{3}{2}}\).

Shortcut (optional).
For a line in the form \(Ax+By+C=0\) with \(B\neq0\),
the slope is \(m=-\dfrac{A}{B}\).

Here \(A=3\), \(B=-2\), so
\(m=-\dfrac{3}{-2}=\dfrac{3}{2}\) again.

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Slope of a line from a standard form equation — Quiz

Each line is written in standard form \(ax + by + c = 0\). The slope is: \[ m = -\frac{a}{b}. \]

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Equation of a Line

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Key Idea — Equation of a Line

A straight-line equation can be written in several equivalent forms:

\( y = mx + c \)
\( ax + by + c = 0 \)
\( y - y_1 = m(x - x_1) \)

Equation of a Line: \(y = mx + c\)

Coordinate Geometry

Explore how the equation of a line is formed from the slope and \(y\)-intercept.

Coordinate Grid Quiz — Equation y=mx+c

Points A and B are shown. Choose the correct equation of the line in the form y=mx+c.

Question

A (y-intercept) B

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Equation of a Perpendicular Line through a Point

Find the equation of the line which passes through \( (-2,1) \) and is perpendicular to the line \( 3x+2y-4=0 \).

Show worked solution

Step 1: Find the slope of the given line.
Rearrange \(3x+2y-4=0\) to slope–intercept form:

\(2y=-3x+4 \Rightarrow y=-\dfrac{3}{2}x+2\).

So the slope of this line is \(m_1=-\dfrac{3}{2}\).

Step 2: Use the fact about perpendicular slopes.
If two lines are perpendicular, then \(m_1 \times m_2=-1\).
So the slope of the required line is

\(m_2=\dfrac{2}{3}\) (the negative reciprocal of \(-\dfrac{3}{2}\)).

Step 3: Use point–slope form with point \( (-2,1) \).
\(y-y_1=m_2(x-x_1)\)

\(y-1=\dfrac{2}{3}\bigl(x-(-2)\bigr) =\dfrac{2}{3}(x+2).\)

Step 4: Simplify to a final equation.
Multiply across by \(3\):
\(3(y-1)=2(x+2)\)
\(3y-3=2x+4\)
\(2x-3y+7=0\).

So the required line is
\(\boxed{2x-3y+7=0}\) (or \(y=\dfrac{2}{3}x+\dfrac{7}{3}\)).

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Using the equation of a line formula

For each question, you are given a point on the line and the slope. Fill in the correct values of \(x_1\), \(y_1\) and \(m\) in:

\(y-y_1 = m(x-x_1)\).

y − = (x − )

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Practice Questions

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Strategies — Solving Coordinate Geometry Problems

• Draw a clear coordinate diagram and label all given points.
• Identify which formula applies: distance, midpoint, slope, or line equation.
• Use the diagram to guide your algebra and check if the answer fits.

Coordinate Geometry of the Line — Practice Questions

Junior Cycle Maths: distance, midpoint, slope, parallel & perpendicular lines, equations, intercepts, and intersection points.

1

Distance between two points

For points \(A(x_1,y_1)\) and \(B(x_2,y_2)\): \[ |AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. \]

Worked example

Find the distance between \(A(1,2)\) and \(B(5,5)\). Show solution

Substitute the coordinates into the distance formula. \(|AB|=\sqrt{(5-1)^2+(5-2)^2}\)

Work out the differences and squares. \(|AB|=\sqrt{4^2+3^2}=\sqrt{16+9}\)

Add and simplify the surd. \(|AB|=\sqrt{25}=5\)

Distance between the points. Therefore \(|AB|=5\) units.

Practice question

Find the distance between \(P(-3,4)\) and \(Q(9,-2)\). Reveal

Use the formula with \(P\) and \(Q\). \(|PQ|=\sqrt{(9-(-3))^2+(-2-4)^2}\)

Evaluate the differences and squares. \(|PQ|=\sqrt{12^2+(-6)^2}=\sqrt{144+36}\)

Simplify the surd. \(|PQ|=\sqrt{180}=6\sqrt{5}\)

2

Midpoint of a line segment

The midpoint of \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is: \[ M\!\left(\tfrac{x_1+x_2}{2},\tfrac{y_1+y_2}{2}\right). \]

Worked example

Find the midpoint of \(A(2,3)\) and \(B(6,7)\). Show solution

Add the x-coordinates and divide by 2. \(x\text{-value}=\tfrac{2+6}{2}=4\)

Add the y-coordinates and divide by 2. \(y\text{-value}=\tfrac{3+7}{2}=5\)

Write the midpoint. \(M=(4,5)\)

Practice question

Find the midpoint of \(C(-5,8)\) and \(D(3,-4)\). Reveal

x-value.\(\tfrac{-5+3}{2}=-1\)

y-value.\(\tfrac{8+(-4)}{2}=2\)

Midpoint.\(M=(-1,2)\)

3

Slope (gradient) of a line

Through \(A(x_1,y_1)\) and \(B(x_2,y_2)\): \[ m=\frac{y_2-y_1}{x_2-x_1}. \]

Worked example

Find the slope through \(A(2,1)\) and \(B(6,5)\). Show solution

Substitute into the formula. \(m=\dfrac{5-1}{6-2}\)

Calculate the differences. \(m=\dfrac{4}{4}\)

Simplify the fraction. \(m=1\)

Practice question

Find the slope through \(P(-3,7)\) and \(Q(9,1)\). Reveal

Apply the formula. \(m=\dfrac{1-7}{9-(-3)}\)

Compute. \(m=\dfrac{-6}{12}\)

Simplify. \(m=-\tfrac{1}{2}\)

4

Parallel and perpendicular lines

Parallel: \(m_1=m_2\). Perpendicular: \(m_1m_2=-1\) (so \(m_2=-\tfrac{1}{m_1}\), \(m_1\neq 0\)).

Worked example

A line has equation \(y=2x+3\). Find slopes of (i) a parallel line and (ii) a perpendicular line. Show solution

Read the slope from \(y=mx+c\).\(m=2\)

Parallel lines share the same slope.Parallel slope \(=2\)

Perpendicular slope is the negative reciprocal.Perpendicular slope \(=-\tfrac{1}{2}\)

Practice question

Find the equation of the line through \((4,-1)\) perpendicular to \(y=\tfrac{3}{2}x+5\). Reveal

Given slope \(m_1=\tfrac{3}{2}\); find the perpendicular slope.\(m_2=-\tfrac{2}{3}\)

Use point–slope with point \((4,-1)\).\(y+1=-\tfrac{2}{3}(x-4)\)

Simplify to slope–intercept form.\(y=-\tfrac{2}{3}x+\tfrac{5}{3}\)

5

Equation of a line

Forms: \[ y=mx+c,\quad y-y_1=m(x-x_1),\quad ax+by+c=0. \]

Worked example

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

Find the slope from the two points.\(m=\dfrac{6-2}{3-1}=2\)

Use point–slope with \((1,2)\).\(y-2=2(x-1)\)

Simplify.\(y=2x\)

Practice question

Find the equation through \((-2,3)\) and \((4,-1)\) in the form \(ax+by+c=0\). Reveal

Slope.\(m=\dfrac{-1-3}{4-(-2)}=-\tfrac{2}{3}\)

Point–slope with \((-2,3)\).\(y-3=-\tfrac{2}{3}(x+2)\)

Simplify to \(y=mx+c\).\(y=-\tfrac{2}{3}x+\tfrac{5}{3}\)

Clear fractions to general form.\(2x+3y-5=0\)

6

x- and y-intercepts

For \(y=mx+c\): y-intercept \(=c\); x-intercept solves \(0=mx+c\Rightarrow x=-\tfrac{c}{m}\) (when \(m\neq0\)).

Worked example

Find the intercepts of \(y=2x+3\). Show solution

Set \(x=0\) for the y-intercept.\(y=3\Rightarrow (0,3)\)

Set \(y=0\) for the x-intercept.\(0=2x+3\Rightarrow x=-\tfrac{3}{2}\Rightarrow (-1.5,0)\)

Practice question

Find the intercepts of \(3x-4y+8=0\). Reveal

y-intercept: put \(x=0\).\(-4y+8=0\Rightarrow y=2\Rightarrow (0,2)\)

x-intercept: put \(y=0\).\(3x+8=0\Rightarrow x=-\tfrac{8}{3}\Rightarrow\left(-\tfrac{8}{3},0\right)\)

7

Point of intersection (elimination method)

Solve two linear equations by making coefficients the same and **eliminating** one variable.

Worked example

Find the intersection of \[ \begin{cases} 2x+y=5\\ x-y=1 \end{cases} \] Show solution

Line up the equations. \(\begin{aligned} &2x+y=5 \quad (1)\\ &x-y=1 \quad (2) \end{aligned}\)

Add (1) and (2) to eliminate \(y\). \((2x+y)+(x-y)=5+1 \Rightarrow 3x=6\)

Solve for \(x\).\(x=2\)

Substitute \(x=2\) into (2) to find \(y\).\(2 - y = 1 \Rightarrow y=1\)

Write the intersection point.\((x,y)=(2,1)\)

Practice question

Find the intersection of \[ \begin{cases} 3x+2y=7\\ x-y=1 \end{cases} \] Reveal

Make the \(x\)-coefficients match by multiplying the second equation by 3. \((x-y=1)\times 3 \Rightarrow 3x-3y=3\)

Subtract the new equation from \(3x+2y=7\) to eliminate \(x\). \((3x+2y)-(3x-3y)=7-3 \Rightarrow 5y=4\)

Solve for \(y\).\(y=\tfrac{4}{5}\)

Back-substitute into \(x-y=1\).\(x-\tfrac{4}{5}=1 \Rightarrow x=\tfrac{9}{5}\)

Intersection point.\(\left(\tfrac{9}{5},\tfrac{4}{5}\right)\)

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Tutorial Videos

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Video Tutorials — Coordinate Geometry

Distance Formula

Slope and Distance

Equation of a Line

Convert Standard Form to Slope–Intercept Form

Intersection of Two Lines

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