Coordinate Geometry of the Line
Basic Concepts Area of Triangle Perpendicular Distance Angle Between Lines Division Triangle Concurrencies Revision and Exam Questions Syllabus

Coordinate Geometry of the Line — Study Checklist

✅ Confident — I can do this independently Need Practice — I get it but need more reps ⛔ Not Yet — I do not understand this yet

Junior Cycle Prior Knowledge — The Line

1. Distance, midpoint and slope

For two points \(A(x_1,y_1)\), \(B(x_2,y_2)\):

\[ d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}, \qquad M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right), \qquad m=\frac{y_2-y_1}{x_2-x_1}. \]

Sample questions
  1. Diagram for distance, midpoint and slope question with A(1,3) and B(5,9) \(A(1,3)\), \(B(5,9)\).
    (a) Find \(|AB|\).
    (b) Find the midpoint.
    (c) Find the slope.
    Distance: \[ |AB|=\sqrt{(5-1)^2+(9-3)^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}. \] Midpoint: \[ M=\left(\frac{1+5}{2},\frac{3+9}{2}\right)=(3,6). \] Slope: \[ m=\frac{9-3}{5-1}=\frac{6}{4}=\frac32. \]




  2. A line passes through \(P(-2,4)\) and \(Q(6,k)\).
    If \(m=\tfrac32\), find \(k\).
    \[ \frac{k-4}{6-(-2)}=\frac32 \quad\Rightarrow\quad \frac{k-4}{8}=\frac32. \] So \[ k-4 = 12 \quad\Rightarrow\quad k = 16. \]

2. Equation forms of a line

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

Sample questions
  1. Find the equation of the line through \(A(2,-1)\) and \(B(6,3)\).
    Slope: \[ m=\frac{3-(-1)}{6-2}=\frac{4}{4}=1. \] Using \(y=mx+c\): Sub in point \(A\): \[ -1=1(2)+c \Rightarrow c=-3. \] Equation: \[ y = x - 3. \]
  2. A line with slope \(-\tfrac12\) passes through \((4,5)\). Write in point–slope form and slope–intercept form.
    Point–slope: \[ y-5=-\tfrac12(x-4). \] Rearranged: \[ y=-\tfrac12 x + 2 + 5 = -\tfrac12 x + 7. \]
  3. Convert \(3x - 2y + 6 = 0\) to \(y = mx + c\).
    \[ -2y = -3x - 6 \quad\Rightarrow\quad y = \frac{3}{2}x + 3. \] Slope \(m=\frac32\), intercept \(c=3\).

3. Parallel and perpendicular lines

Parallel: \(m_1=m_2\)     Perpendicular: \(m_1m_2=-1\)

Sample questions
  1. Are the lines parallel, perpendicular or neither? \(y=3x+4\), \(y-2=3(x+1)\)
    Second line rearranges to \(y = 3x + 5\). Both have slope \(3\) ⇒ **parallel**.
  2. Find the equation of the line parallel to \(3x-2y+5=0\) through \((1,-4)\).
    Rearrange original: \[ -2y = -3x - 5 \Rightarrow y = \tfrac32 x + \tfrac52. \] Use slope \(m=\frac32\): Through \((1,-4)\): \[ y+4 = \tfrac32(x-1) \] \[ y = \tfrac32 x - \tfrac32 - 4 = \tfrac32 x - \tfrac{11}{2}. \]
  3. Find the equation perpendicular to \(y=-\tfrac34x+2\) through \((2,1)\).
    Perpendicular slope: \[ m_\perp = \frac{4}{3}. \] Equation: \[ y-1 = \frac43(x-2) \] \[ y = \frac43x - \frac83 + 1 = \frac43x - \frac53. \]

4. Intersection of lines, rate of change, intercept

Sample questions
  1. Find the intersection of \(y = 2x + 1\) and \(y = -x + 7\).
    Solve: \[ 2x+1 = -x+7 \] \[ 3x = 6 \Rightarrow x = 2. \] Then \[ y = 2(2)+1 = 5. \] Intersection: \((2,5)\).
  2. Taxi fare: \(\text{Cost} = 3x + 5\). Identify the initial cost and the cost per km.
    Fixed charge \(=5\) euro. Rate of change \(=3\) euro per kilometre.
  3. Cooling model: \(T = 20 - 4t\).
    At \(t=0\): \(T = 20^\circ\text{C}\). The slope \(-4\) means the drink cools at \(4^\circ\text{C}\) per minute.

Interactive: Area by Translating \(A\) to \((0,0)\)

\[ \text{Area} = \frac{1}{2}\left|x_1y_2 - x_2y_1\right|. \]
Enter coordinates for \(A,B,C\). The triangle is translated so that \(A\) is at the origin, and the worked solution shows the calculation of the area.

x₁ = y₁ =
x₂ = y₂ =
x₃ = y₃ =
Original triangle Translated triangle

Area of a Triangle by Translating \(A\) to the Origin

To find the area of a triangle in coordinate geometry, we can simplify the calculation by translating one vertex to the origin. Here we translate \(A\) to \((0,0)\).

Area formula when a vertex is at the origin
If the triangle is \[ O(0,0),\quad P(x_1,y_1),\quad Q(x_2,y_2), \] then \[ \text{Area} = \frac{1}{2}\left|x_1y_2 - x_2y_1\right|. \]
Translation rule (move \(A\) to \((0,0)\))
For a triangle with vertices \[ A(x_1,y_1),\quad B(x_2,y_2),\quad C(x_3,y_3), \] translate by subtracting the coordinates of \(A\) from every point: \[ A'(0,0),\qquad B'(x_2-x_1,\;y_2-y_1),\qquad C'(x_3-x_1,\;y_3-y_1). \]

Sample question

Find the area of the triangle with vertices \[ A(2,1),\quad B(8,4),\quad C(5,7). \]

Step 1: Translate the triangle so that \(A\) becomes \((0,0)\).

Subtract the coordinates of \(A(2,1)\) from each vertex: \[ A'(0,0),\qquad B'(8-2,\;4-1) = (6,3),\qquad C'(5-2,\;7-1) = (3,6). \]

Step 2: Substitute directly into the area formula.

Now the triangle is \[ O(0,0),\quad P(6,3),\quad Q(3,6). \] Use \[ \text{Area}=\frac{1}{2}\lvert x_1y_2 - x_2y_1\rvert. \] Substitute: \[ \text{Area} =\frac{1}{2}\,\lvert 6\cdot 6 - 3\cdot 3\rvert =\frac{1}{2}\,\lvert 36 - 9\rvert =\frac{1}{2}\cdot 27 =13.5. \]

The area of the triangle is \[ \boxed{13.5\ \text{square units}}. \]

Practice — Area of a Triangle (Translate \(A\) to \((0,0)\))

For each question, first translate \(A\) so that it becomes \((0,0)\), then use \[ \text{Area}=\frac{1}{2}\bigl|x_1y_2 - x_2y_1\bigr| \] with the translated coordinates.

  1. Find the area of the triangle with vertices \[ A(1,2),\quad B(7,5),\quad C(4,9). \]

    Translate \(A\) to the origin.

    Subtract \(A(1,2)\) from each point: \[ A'(0,0),\quad B'(7-1,\;5-2)=(6,3),\quad C'(4-1,\;9-2)=(3,7). \]

    Use the area formula.

    Take \(P=B'(6,3)\), \(Q=C'(3,7)\). Then \[ \text{Area} =\frac{1}{2}\lvert 6\cdot7-3\cdot3\rvert =\frac{1}{2}\lvert42-9\rvert =\frac{1}{2}\cdot33 =16.5. \]

    So the area is \[ \boxed{16.5\ \text{square units}}. \]

  2. Find the area of the triangle with vertices \[ A(-2,1),\quad B(3,4),\quad C(5,-1). \]

    Translate \(A\) to the origin.

    Subtract \(A(-2,1)\) from each point: \[ A'(0,0),\quad B'(3-(-2),\;4-1)=(5,3),\quad C'(5-(-2),\;-1-1)=(7,-2). \]

    Use the area formula.

    Take \(P=B'(5,3)\), \(Q=C'(7,-2)\). Then \[ \text{Area} =\frac{1}{2}\lvert 5(-2)-7\cdot3\rvert =\frac{1}{2}\lvert -10-21\rvert =\frac{1}{2}\cdot31 =15.5. \]

    So the area is \[ \boxed{15.5\ \text{square units}}. \]

  3. The triangle has vertices \[ A(1,2),\quad B(7,5),\quad C(x,y). \] Its area is \(16.5\) square units. Find one possible set of integer coordinates for the vertex \(C\).

    Translate \(A\) to the origin.

    Subtract \(A(1,2)\) from each point: \[ A'(0,0),\quad B'(7-1,\;5-2)=(6,3),\quad C'(x-1,\;y-2). \]

    Use the area formula with an unknown point.

    Let \(P=B'(6,3)\), \(Q=C'(x-1,y-2)\). Then \[ \text{Area} =\frac{1}{2}\left|6(y-2)-3(x-1)\right|. \] We are told the area is \(16.5\), so \[ \frac{1}{2}\bigl|6(y-2)-3(x-1)\bigr| = 16.5 \quad\Rightarrow\quad \bigl|6(y-2)-3(x-1)\bigr| = 33. \]

    One way is to choose integers so that \[ 6(y-2)-3(x-1)=33. \] Divide by \(3\): \[ 2(y-2)-(x-1)=11 \quad\Rightarrow\quad 2y-4-x+1=11 \quad\Rightarrow\quad 2y-x=14. \]

    Choose a convenient integer value for \(x\), for example \(x=2\): \[ 2y-2=14 \quad\Rightarrow\quad 2y=16 \quad\Rightarrow\quad y=8. \] So one possible translated point is \[ C'(x-1,y-2) = (2-1,\;8-2) = (1,6). \]

    Translate back by adding the coordinates of \(A(1,2)\): \[ C(x,y) = (1+1,\;2+6) = (2,8). \]

    Check: \[ \text{Area} =\frac{1}{2}\bigl|6\cdot6-3\cdot3\bigr| =\frac{1}{2}\bigl|36-9\bigr| =\frac{1}{2}\cdot27 =16.5. \]

    So one possible vertex is \[ \boxed{C(2,8)}. \] (There are infinitely many possible points \(C\); this is just one integer example.)

Distance from a Point to a Line

In the coordinate plane, the distance from a point \(P(x_1,y_1)\) to a line \[ ax + by + c = 0 \] is the length of the perpendicular from \(P\) to the line.

Formula
\[ d = \frac{\lvert a x_1 + b y_1 + c\rvert}{\sqrt{a^2 + b^2}}. \]
Sample Question
Find the perpendicular distance from the point \(P(4,-1)\) to the line \[ 3x - 4y - 12 = 0. \]

Substitute \(a = 3\), \(b = -4\), \(c = -12\), \(x_1 = 4\), \(y_1 = -1\) directly into the formula:

\[ d = \frac{\lvert 3(4) + (-4)(-1) - 12\rvert}{\sqrt{3^2 + (-4)^2}} = \frac{\lvert 12 + 4 - 12\rvert}{\sqrt{9 + 16}} = \frac{\lvert 4\rvert}{\sqrt{25}} = \frac{4}{5}. \]

Therefore the perpendicular distance is \[ \boxed{\frac{4}{5} = 0.8}. \]

Interactive: Distance from a Point to a Line

Enter a line \(ax + by + c = 0\) and a point \(P(x_0,y_0)\). The diagram and worked solution update instantly.

a = b = c =
x₀ = y₀ =
Distance formula
\[ d=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}. \]

Distance from a Point to a Line — Practice

Use \(d = \dfrac{\lvert a x_1 + b y_1 + c\rvert}{\sqrt{a^2 + b^2}}\) to find the perpendicular distance in each case.

  1. Find the perpendicular distance from the point \(P(1,2)\) to the line \[ 2x - 3y + 6 = 0. \]
    Here \(a = 2\), \(b = -3\), \(c = 6\), \(x_1 = 1\), \(y_1 = 2\). \[ a x_1 + b y_1 + c = 2(1) + (-3)(2) + 6 = 2 - 6 + 6 = 2. \] \[ \sqrt{a^2 + b^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}. \] So \[ d = \frac{\lvert 2\rvert}{\sqrt{13}} = \frac{2}{\sqrt{13}} \approx 0.555. \]
  2. The line \(L\) has equation \[ 4x + 3y - 12 = 0. \] Find the distance from the origin to the line \(L\).
    The origin is \(O(0,0)\), so \(x_1 = 0\), \(y_1 = 0\). \[ a x_1 + b y_1 + c = 4(0) + 3(0) - 12 = -12, \] so \[ \lvert a x_1 + b y_1 + c\rvert = 12. \] \[ \sqrt{a^2 + b^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5. \] Therefore \[ d = \frac{12}{5} = 2.4. \]
  3. The point \(A(5,-3)\) lies above the line \[ x - 2y - 1 = 0. \] Show that the distance from \(A\) to the line is \(\sqrt{5}\).
    Here \(a = 1\), \(b = -2\), \(c = -1\), \(x_1 = 5\), \(y_1 = -3\). \[ a x_1 + b y_1 + c = 1(5) + (-2)(-3) - 1 = 5 + 6 - 1 = 10. \] \[ \sqrt{a^2 + b^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}. \] Hence \[ d = \frac{\lvert 10\rvert}{\sqrt{5}} = \frac{10}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}. \] So the distance is \(2\sqrt{5}\), not \(\sqrt{5}\); the statement in the question is false.
  4. The line \[ 3x + 4y + k = 0 \] is at a distance \(5\) units from the origin. Find the possible values of \(k\).
    For the origin \(O(0,0)\) we have \[ d = \frac{\lvert 3\cdot 0 + 4\cdot 0 + k\rvert}{\sqrt{3^2 + 4^2}} = \frac{\lvert k\rvert}{5}. \] We are told that \(d = 5\), so \[ \frac{\lvert k\rvert}{5} = 5 \quad\Rightarrow\quad \lvert k\rvert = 25. \] Therefore \[ k = 25 \quad\text{or}\quad k = -25. \]

Dividing a Line Segment in a Given Ratio

Let \(A(x_1,y_1)\) and \(B(x_2,y_2)\) be two points in the plane. A point \(P\) lies on the line segment \(AB\) and divides it internally in the ratio \(m:n\) if \[ AP:PB = m:n,\qquad m,n > 0. \]

Section formula (internal division)
If \(P\) divides the segment from \(A(x_1,y_1)\) to \(B(x_2,y_2)\) internally in the ratio \(m:n\) (with \(AP:PB = m:n\)), then \[ P\left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right). \] This is a weighted average of the coordinates of \(A\) and \(B\). When \(m = n\), the point \(P\) is the midpoint of \(AB\).
Sample question
The points \(A(-2,3)\) and \(B(7,9)\) are the ends of a line segment. Find the coordinates of the point \(P\) that divides the segment internally in the ratio \(AP:PB = 2:3\).
Here \(m = 2\), \(n = 3\), \(x_1 = -2\), \(y_1 = 3\), \(x_2 = 7\), \(y_2 = 9\). Using the section formula, \[ x_P = \frac{m x_2 + n x_1}{m+n} = \frac{2(7) + 3(-2)}{2+3} = \frac{14 - 6}{5} = \frac{8}{5}, \] \[ y_P = \frac{m y_2 + n y_1}{m+n} = \frac{2(9) + 3(3)}{5} = \frac{18 + 9}{5} = \frac{27}{5}. \] Therefore \[ P\left(\frac{8}{5},\;\frac{27}{5}\right). \]

Interactive: Dividing a Line Segment in the Ratio m : n

Choose the end points A(x1, y1) and B(x2, y2), then use the slider to set the ratio AP : PB = m : n.

Current ratio: m = 3, n = 7
Point A
Point B
Point P
Section formula (internal)
P ( (m x2 + n x1)/(m + n), (m y2 + n y1)/(m + n) )
Using A(x1, y1) = (, ), B(x2, y2) = (, ), m = , n = :
P = \( \big( (m x_2 + n x_1)/(m+n), (m y_2 + n y_1)/(m+n) \big) \) = \( \Big( \dfrac{}{}, \dfrac{}{} \Big) \).
Current point: P( , )

Practice: Dividing a Line Segment in a Given Ratio

Find the coordinates of the point \(P\) that divides the line segment internally.

  1. \(A(2,-1)\), \(B(8,5)\), ratio \(1:2\).
    \[ P\left(\frac{1\cdot 8 + 2\cdot 2}{3}, \frac{1\cdot 5 + 2\cdot(-1)}{3}\right) = (4,1). \]
  2. \(A(-3,4)\), \(B(9,1)\), ratio \(3:1\).
    \[ P=\left(\frac{3\cdot 9 + 1\cdot(-3)}{4}, \frac{3\cdot1 + 1\cdot 4}{4}\right) = (6,\tfrac{7}{4}). \]
  3. \(A(-5,-2)\), \(B(1,10)\), ratio \(2:5\).
    \[ P=\left(\frac{2\cdot1 + 5\cdot(-5)}{7}, \frac{2\cdot10 + 5\cdot(-2)}{7}\right) =\left(\!-\frac{23}{7}, \frac{10}{7}\right). \]
  4. \(A(4,-3)\), \(B(-2,9)\), ratio \(5:1\).
    \[ P = \left(\frac{5(-2)+4}{6}, \frac{5\cdot 9 -3}{6}\right) = (-1,7). \]
  5. \(P(1,2)\) divides the segment from \(A(-5,8)\) to \(B(x,-4)\) in the ratio \(1:3\). Find \(x\).
    \[ 1=\frac{x - 15}{4}\quad\Rightarrow\quad x=19. \]
  6. \(P(3,k)\) divides the segment from \(A(-1,2)\) to \(B(7,-6)\) in the ratio \(3:1\). Find \(k\).
    \[ k = \frac{3(-6)+2}{4} = -4. \]

Interactive: Angle Between Two Lines

Adjust the slopes of two lines through the origin. The diagram and the acute angle update instantly.

1.0
-2.0
Angle formula:
tan θ = |(m₂ − m₁) / (1 + m₁m₂)|
Line 1
Line 2
Angle arc
Note: If m₁·m₂ = −1, the lines are perpendicular.
m₁ = , m₂ = , θ ≈ °

Angle Between Two Lines

Consider two non-vertical lines with slopes \(m_1\) and \(m_2\). Let \(\theta\) be the (acute) angle between them.

Formula using slopes
The angle \(\theta\) between the lines is given by \[ \tan\theta = \left|\,\frac{m_2 - m_1}{1 + m_1 m_2}\,\right|. \] We usually find \(\theta\) by taking \[ \theta = \tan^{-1}\!\left(\left|\,\frac{m_2 - m_1}{1 + m_1 m_2}\,\right|\right), \] and choosing the acute angle (between \(0^\circ\) and \(90^\circ\)).

If the lines are given in the form \[ L_1: a_1 x + b_1 y + c_1 = 0,\qquad L_2: a_2 x + b_2 y + c_2 = 0, \] then their slopes are \[ m_1 = -\frac{a_1}{b_1},\qquad m_2 = -\frac{a_2}{b_2}, \] provided \(b_1\) and \(b_2\) are not zero.

Special cases

  • If \(m_1 m_2 = -1\), then \(\theta = 90^\circ\) (lines perpendicular).
  • If \(m_1 = m_2\), then \(\theta = 0^\circ\) (lines parallel).
Sample question
Find the acute angle between the lines \[ L_1: 3x - 4y + 5 = 0, \qquad L_2: x + 2y - 7 = 0. \]

Step 1: Find the slopes.

For \(L_1: 3x - 4y + 5 = 0\), \[ y = \frac{3}{4}x + \frac{5}{4}, \] so \(m_1 = \tfrac{3}{4}\).

For \(L_2: x + 2y - 7 = 0\), \[ y = -\frac{1}{2}x + \frac{7}{2}, \] so \(m_2 = -\tfrac{1}{2}\).

Step 2: Use the angle formula.

\[ \tan\theta = \left|\,\frac{-\tfrac{1}{2} - \tfrac{3}{4}}{1 + \tfrac{3}{4}(-\tfrac{1}{2})}\,\right| = 2. \] \[ \theta = \tan^{-1}(2) \approx 63.4^\circ. \]

Therefore the acute angle between the lines is \[ \boxed{63.4^\circ}. \]

Angle Between Two Lines — Practice

Find the **acute angle** between each pair of lines.

  1. Find the angle between \[ L_1: y = 2x + 3,\qquad L_2: y = -x + 5. \]
    Slopes: \(m_1 = 2\), \(m_2 = -1\). \[ \tan\theta = \left|\frac{m_2 - m_1}{1 + m_1m_2}\right| = \left|\frac{-1 - 2}{1 + 2(-1)}\right| = |-3/-1| = 3. \] \[ \theta = \tan^{-1}(3) \approx 71.6^\circ. \]
  2. Find the angle between \[ L_1: 2x + 3y - 7 = 0,\qquad L_2: 3x - 4y + 1 = 0. \]
    From rearranging, \[ m_1 = -\frac{2}{3},\qquad m_2 = \frac{3}{4}. \] \[ \tan\theta = \left|\frac{\tfrac34 - \left(-\tfrac23\right)}{1 + \left(-\tfrac23\right)\left(\tfrac34\right)}\right| = \frac{17/12}{1/2} = \frac{17}{6}. \] \[ \theta = \tan^{-1}\!\left(\frac{17}{6}\right) \approx 70.5^\circ. \]
  3. Show that \[ L_1: y = \tfrac12 x + 1,\qquad L_2: y = -2x + 3 \] are perpendicular. Hence state the angle between them.
    \[ m_1 = \frac12,\qquad m_2 = -2. \] \[ m_1m_2 = -1. \] Therefore the lines are perpendicular. \[ \theta = 90^\circ. \]
  4. The lines \[ L_1: y = 3x + 1,\qquad L_2: y = kx - 5 \] meet at an acute angle of \(45^\circ\). Find \(k\).
    \[ \left|\frac{k - 3}{1 + 3k}\right| = 1. \] Case 1: \[ \frac{k - 3}{1 + 3k} = 1 \Rightarrow k = -2. \] Case 2: \[ \frac{k - 3}{1 + 3k} = -1 \Rightarrow k = \tfrac12. \] \[ \boxed{k = -2 \text{ or } k = \tfrac12.} \]

Interactive: Triangle Concurrencies

Explore how the four key triangle centres — the circumcentre, incentre, centroid, and orthocentre — behave as you move the vertices of the triangle.

  • Understand how each centre of a triangle is constructed.
  • Observe the relationships between medians, angle bisectors, perpendicular bisectors, and altitudes.
  • Investigate how each centre behaves for acute, obtuse, and right-angled triangles.

Triangle Concurrencies

Select a Polypad to display the geometric or coordinate-geometric construction below.

Polypad view

Triangle Points of Concurrency

Click a card to reveal the definition, key facts, or construction steps.

Key ideas and properties

Definition: The incentre is the point where the three internal angle bisectors meet.

  • Equidistant from all three sides.
  • Centre of the inscribed circle (incircle).

Definition: The circumcentre is the point where the three perpendicular bisectors of the sides meet.

  • Equidistant from all three vertices.
  • Centre of the circumcircle.

Definition: The centroid is where the three medians meet.

  • Always inside the triangle.
  • Each median is divided in the ratio \(2:1\) (vertex to centroid : centroid to midpoint).

Definition: The orthocentre is the point where the three altitudes meet.

  • Inside for an acute triangle, at the right angle for a right-angled triangle, outside for an obtuse triangle.
Compass and ruler constructions

Steps:

  • Construct the angle bisector at vertex \(A\).
  • Construct the angle bisector at vertex \(B\).
  • The two bisectors meet at the incentre \(I\).
  • Use \(I\) and the perpendicular distance to a side as radius to draw the incircle.

Steps:

  • Construct the perpendicular bisector of \(AB\).
  • Construct the perpendicular bisector of \(AC\).
  • The bisectors meet at the circumcentre \(O\).
  • Use radius \(OA\) (or \(OB\) or \(OC\)) to draw the circumcircle.

Steps:

  • Find the midpoint of \(BC\).
  • Join \(A\) to this midpoint (a median).
  • Repeat for another side (e.g. median from \(B\)).
  • The medians meet at the centroid \(G\).

Steps:

  • Construct the altitude from \(A\) to \(BC\).
  • Construct the altitude from \(B\) to \(AC\).
  • The altitudes meet at the orthocentre \(H\) (extend sides if the triangle is obtuse).

Centroid of a Triangle

Consider a triangle with vertices \[ A(x_1,y_1),\quad B(x_2,y_2),\quad C(x_3,y_3). \] The centroid \(G\) is the point where the three medians meet. (A median joins a vertex to the midpoint of the opposite side.)

Centroid formula
The centroid \(G\) of triangle \(ABC\) is the “average” of the three vertices: \[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right). \] Each coordinate of \(G\) is the mean of the corresponding coordinates of the three vertices.

Worked example

Example
Find the coordinates of the centroid of the triangle with vertices \[ A(1,4),\quad B(7,1),\quad C(4,7). \]

Using the centroid formula \[ G\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right), \] we substitute \[ x_1 = 1,\quad x_2 = 7,\quad x_3 = 4, \qquad y_1 = 4,\quad y_2 = 1,\quad y_3 = 7. \]

\[ x_G = \frac{1 + 7 + 4}{3} = \frac{12}{3} = 4, \qquad y_G = \frac{4 + 1 + 7}{3} = \frac{12}{3} = 4. \]

Therefore the centroid is \[ G(4,4). \]

Triangle Concurrencies — Centroid, Orthocentre, Circumcentre

  1. The vertices of a triangle are \[ A(2,1),\quad B(8,4),\quad C(5,7). \] (a) Find the centroid \(G\). (b) Show that \(G\) lies on the median from \(A\).
    (a) Centroid
    \[ G\left(\frac{2+8+5}{3},\frac{1+4+7}{3}\right) = (5,4). \] (b) Check median
    Midpoint of \(BC\): \[ M\left(\frac{8+5}{2},\frac{4+7}{2}\right) = \left(\frac{13}{2}, \frac{11}{2}\right). \] Slopes: \[ m_{AM}=1,\quad m_{AG}=1. \] So \(A\), \(G\), \(M\) are collinear → \(G\) lies on the median.
  2. The triangle \(PQR\) has vertices \[ P(1,1),\quad Q(7,2),\quad R(3,8). \] Find the coordinates of the orthocentre \(H\).

    Slopes of the sides

    \[ m_{PQ} = \frac{2-1}{7-1} = \frac{1}{6},\quad m_{PR} = \frac{8-1}{3-1} = \frac{7}{2},\quad m_{QR} = \frac{8-2}{3-7} = -\frac{3}{2}. \]

    Altitude from \(R\)

    Perpendicular to \(PQ\): slope \(-6\). Through \(R(3,8)\): \[ y - 8 = -6(x - 3)\;\Rightarrow\; y = -6x + 26. \]

    Altitude from \(P\)

    Perpendicular to \(QR\): slope \(+\frac{2}{3}\). Through \(P(1,1)\): \[ y - 1 = \frac{2}{3}(x - 1)\;\Rightarrow\; y = \frac{2}{3}x + \frac{1}{3}. \]

    Intersect the altitudes

    Solve \[ -6x + 26 = \frac{2}{3}x + \frac{1}{3} \] \[ -\frac{20}{3}x = -\frac{77}{3} \Rightarrow x = \frac{77}{20}. \] Then \[ y = \frac{2}{3}\cdot\frac{77}{20} + \frac{1}{3} = \frac{29}{10}. \] So the orthocentre is \[ H\left(\frac{77}{20},\;\frac{29}{10}\right)\approx(3.85,2.90). \]
  3. The vertices of a triangle are \[ A(1,2),\quad B(7,3),\quad C(4,8). \] Find the coordinates of the circumcentre \(O\).

    Midpoints

    \[ M_{AB}=\left(\frac{1+7}{2},\frac{2+3}{2}\right)=(4,2.5),\quad M_{AC}=\left(\frac{1+4}{2},\frac{2+8}{2}\right)=(2.5,5). \]

    Slopes and perpendicular bisectors

    \[ m_{AB} = \frac{3-2}{7-1} = \frac{1}{6} \Rightarrow m_{\perp AB}=-6, \] \[ m_{AC} = \frac{8-2}{4-1} = 2 \Rightarrow m_{\perp AC}=-\frac{1}{2}. \] Perpendicular bisectors: \[ \text{Through }M_{AB}(4,2.5):\quad y-2.5=-6(x-4)\Rightarrow y=-6x+26.5, \] \[ \text{Through }M_{AC}(2.5,5):\quad y-5=-\frac12(x-2.5)\Rightarrow y=-\frac12x+7.5. \]

    Intersection → Circumcentre

    \[ -6x+26.5 = -\frac12x+7.5 \Rightarrow -5.5x = -19 \Rightarrow x = \frac{38}{11}. \] \[ y = -\frac12\cdot\frac{38}{11}+7.5 = \frac{127}{22}. \] So \[ O\left(\frac{38}{11},\;\frac{127}{22}\right)\approx(3.46,5.77). \]

The Line — 8 Revision Questions

1. Basic concepts — parallelogram

\(A(2,3)\) and \(B(5,9)\) are two of the vertices of the parallelogram \(ABCD\), and the diagonals intersect at \(P(6,7)\). Find the co-ordinates of \(C\) and \(D\).

In a parallelogram the diagonals bisect each other, so \(P\) is the midpoint of both \(AC\) and \(BD\). Use the midpoint formula \[ M\left(\tfrac{x_1+x_2}{2},\,\tfrac{y_1+y_2}{2}\right). \]
For \(AC\), midpoint \(P(6,7)\) gives \[ \frac{2+x_C}{2}=6,\quad \frac{3+y_C}{2}=7 \] so \(x_C=10,\ y_C=11\), hence \(C(10,11)\).

For \(BD\), \[ \frac{5+x_D}{2}=6,\quad \frac{9+y_D}{2}=7 \] so \(x_D=7,\ y_D=5\), hence \(D(7,5)\).
2. Area of a triangle

The area of the triangle \(OAB\), where \(O=(0,0)\), \(A=(x,6)\) and \(B=(6,2)\), is \(14\) square units. Find the possible values of \(x\).

For a triangle with one vertex at the origin, \[ A=\tfrac12\left|x_1y_2-x_2y_1\right|. \] Take \(A=(x,6)\), \(B=(6,2)\).
Using \[ A=\tfrac12\left|x_1y_2-x_2y_1\right| \] with \(x_1=x,\ y_1=6,\ x_2=6,\ y_2=2\), \[ 14=\tfrac12|x\cdot2-6\cdot6| =\tfrac12|2x-36|. \] Hence \(|2x-36|=28\), so \[ 2x-36=28\quad\text{or}\quad 2x-36=-28. \] Thus \(2x=64\) or \(2x=8\), giving \[ x=32\quad\text{or}\quad x=4. \]
3. Angle between two lines

Find, correct to the nearest degree, the larger angle between the lines \(2x-3y=7\) and \(3x+5y+9=0\).

First find the slopes: write each line as \(y=mx+c\) to get \(m_1\) and \(m_2\). Then use \[ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| \] for the acute angle, and subtract from \(180^\circ\) to get the larger angle.
For \(2x-3y=7\): \(-3y=-2x+7\Rightarrow y=\tfrac23x-\tfrac73\), so \(m_1=\tfrac23\).
For \(3x+5y+9=0\): \(5y=-3x-9\Rightarrow y=-\tfrac35x-\tfrac95\), so \(m_2=-\tfrac35\).

Then \[ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| =\left|\frac{\tfrac23-(-\tfrac35)}{1+\tfrac23\cdot(-\tfrac35)}\right| =\left|\frac{19/15}{3/5}\right| =\frac{19}{9}. \] So \(\theta\approx\arctan\!\left(\tfrac{19}{9}\right)\approx64.7^\circ\).
The larger angle is \(180^\circ-64.7^\circ\approx115^\circ\).
4. Perpendicular distance — parallel lines

The perpendicular distance between the parallel lines \(x+2y=8\) and \(2x+4y+k=0\) is \(\sqrt{5}\). Find the two possible values of \(k\in\mathbb{R}\).

Put both lines in the form \(ax+by+c=0\) with the same \(a\) and \(b\), e.g. \(2x+4y-16=0\) and \(2x+4y+k=0\). For parallel lines \[ d=\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}. \]
First line: \(x+2y=8\Rightarrow 2x+4y-16=0\).
Second line: \(2x+4y+k=0\).

Distance between them: \[ d=\frac{|(-16)-k|}{\sqrt{2^2+4^2}} =\frac{|k+16|}{\sqrt{20}} =\frac{|k+16|}{2\sqrt5}. \] Given \(d=\sqrt5\), \[ \frac{|k+16|}{2\sqrt5}=\sqrt5 \Rightarrow |k+16|=10. \] So \[ k+16=10\ \text{or}\ k+16=-10 \Rightarrow k=-6\ \text{or}\ k=-26. \]
5. Basic concepts — fixed distance on a vertical line

A metal rod has length \(\sqrt{61}\). One end is welded to the point \(A(1,-2)\). The other end is to be welded to a point \(B\) on the vertical line \(x=6\). By letting \(B=(6,k)\), find the co-ordinates of the two possible points \(B\).

Use the distance formula between \(A(1,-2)\) and \(B(6,k)\): \[ AB=\sqrt{(6-1)^2+(k+2)^2}. \] Set this equal to \(\sqrt{61}\) and solve for \(k\).
Distance squared: \[ AB^2=(6-1)^2+(k+2)^2=25+(k+2)^2. \] Given \(AB=\sqrt{61}\), \[ 25+(k+2)^2=61\Rightarrow (k+2)^2=36. \] Thus \(k+2=6\) or \(k+2=-6\), so \[ k=4\quad\text{or}\quad k=-8. \] Hence \(B(6,4)\) or \(B(6,-8)\).
6. Perpendicular distance — point to line

A ship is travelling along the line \(5x-y=18\). A lighthouse is located at the point \((2,4)\). Find, correct to two decimal places, the closest that the ship gets to the lighthouse.

Put the line in the form \(ax+by+c=0\): \(5x-y-18=0\). For the distance from a point \((x_1,y_1)\) to this line, use \[ d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}. \]
Here \(a=5,\ b=-1,\ c=-18,\ (x_1,y_1)=(2,4)\). Then \[ d=\frac{|5\cdot2+(-1)\cdot4-18|}{\sqrt{25+1}} =\frac{|10-4-18|}{\sqrt{26}} =\frac{12}{\sqrt{26}} \approx 2.35. \] So the closest distance is approximately \(2.35\) units.
7. Divisor of a line segment

If \(A=(2,8)\) and \(B=(-8,13)\), find the co-ordinates of the point which divides \([AB]\) in the ratio \(3:2\).

If a point \(P(x,y)\) divides the segment from \(A(x_1,y_1)\) to \(B(x_2,y_2)\) internally in the ratio \(m:n\) (i.e. \(AP:PB=m:n\)), then \[ P\left(\frac{nx_1+mx_2}{m+n},\,\frac{ny_1+my_2}{m+n}\right). \] Here the ratio is \(3:2\) from \(A\) to \(B\).
For \(A(2,8)\), \(B(-8,13)\) and ratio \(3:2\) from \(A\) to \(B\), take \(m=3,\ n=2\). Then \[ x=\frac{2\cdot2+3\cdot(-8)}{3+2} =\frac{4-24}{5} =-4, \] \[ y=\frac{2\cdot8+3\cdot13}{3+2} =\frac{16+39}{5} =11. \] Hence the point which divides \([AB]\) in the ratio \(3:2\) is \[ P(-4,11). \]

HL 2025 – Question 1 (Lines & Coordinate Geometry)

30 marks
▸

(a) Point on a line

\(p \in \mathbb{R}\) is a constant. The point \((p,5)\) lies on the line \[ 3x - 2y + 28 = 0. \] Find the value of \(p\).

Solution (a)

Substitute \((x,y) = (p,5)\) into the line: \[ 3p - 2(5) + 28 = 0. \] \[ 3p - 10 + 28 = 0 \Rightarrow 3p + 18 = 0 \Rightarrow 3p = -18 \Rightarrow p = -6. \]

Marking scheme — Scale 5C (0, 2, 3, 5)

Full credit (5): obtains \(p=-6\) (accept correct answer without work).

High partial (3): relevant work to isolate \(p\), e.g. reaches \(3p+18=0\) but makes an error solving.

Low partial (2): work of merit, e.g. substitutes \(p\) for \(x\) or \(5\) for \(y\) correctly, or finds \(3p-10+28\).

Accept \(x\) used in place of \(p\). If only \((-6,5)\) is substituted and fully verified, award at most high partial credit.

(b) Angle between two lines

The line \(\ell\) has equation \[ y = -\tfrac13 x + 11. \] The line \(h\) has equation \[ 2x - 5y + 10 = 0. \] Work out the size of the acute angle between the lines \(\ell\) and \(h\). Give your answer correct to the nearest degree.

Solution (b)

Slope of \(\ell\): \(m_{\ell} = -\tfrac13.\)

Rearrange \(h\): \[ 2x - 5y + 10 = 0 \Rightarrow -5y = -2x - 10 \Rightarrow y = \tfrac25 x + 2, \] so \(m_h = \tfrac25.\)

For the angle \(\theta\) between two lines, \[ \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|. \] Here \[ \tan\theta = \left| \frac{-\tfrac13 - \tfrac25}{1 + (-\tfrac13)(\tfrac25)} \right| = \frac{11}{13}. \]

\[ \theta = \tan^{-1}\!\left(\frac{11}{13}\right) \approx 40.24^\circ. \] The acute angle between the lines is \[ 40^\circ \] correct to the nearest degree.

Marking scheme — Scale 10D (0, 3, 5, 7, 10)

Consider the solution as 4 steps:

Step 1: Finds \(m_{\ell}\).
Step 2: Finds \(m_h\).
Step 3: Substitutes in the appropriate formula for \(\tan\theta\).
Step 4: Finds \(\theta\).

Full credit (10): all four steps correct, answer \(40^\circ\) (accept correct answer without unit; apply * for incorrect rounding).

High partial (7): three steps correct.

Mid partial (5): two steps correct.

Low partial (3): one step correct, or some correct work towards rearranging the equation of \(h\), or a clear diagram showing a graph of the two lines.

(c) Line cutting both axes

A line cuts the \(x\)-axis at the point \(A(a,0)\) and the \(y\)-axis at \(B(0,b)\), where \(a,b \in \mathbb{Z}\). The slope of this line is \(-\tfrac{2}{3}\). The area of the triangle enclosed by this line, the \(x\)-axis, and the \(y\)-axis is \(12\) square units.

There are two different lines that satisfy these conditions. Find the equation of each of these lines.

Solution (c)

First set of possible lines.

Solution image for Q1(c), method 1

Intercepts \(A(a,0)\) and \(B(0,b)\) give slope \[ m = \frac{b-0}{0-a} = \frac{b}{-a} = -\frac{b}{a}. \] Given \(m = -\tfrac{2}{3}\), \[ -\frac{b}{a} = -\frac{2}{3} \Rightarrow \frac{b}{a} = \frac{2}{3} \Rightarrow 3b = 2a. \]

The area of triangle \(OAB\) is \[ \text{Area} = \tfrac12 ab = 12 \Rightarrow ab = 24. \]

From \(3b = 2a\), we have \(a = \tfrac32 b\). Substitute into \(ab = 24\): \[ \left(\tfrac32 b\right)b = 24 \Rightarrow \tfrac32 b^2 = 24 \Rightarrow b^2 = 16 \Rightarrow b = \pm4. \] Then \(a = \tfrac32 b\) gives \[ (a,b) = (6,4)\quad\text{or}\quad (a,b) = (-6,-4). \]

For intercepts \((6,0)\) and \((0,4)\): \[ \frac{x}{6} + \frac{y}{4} = 1 \Rightarrow 2x + 3y = 12 \Rightarrow y = -\tfrac23 x + 4. \] For intercepts \((-6,0)\) and \((0,-4)\): \[ 2x + 3y = -12 \Rightarrow y = -\tfrac23 x - 4. \]

So the equations of the two lines are \[ y = -\tfrac23 x + 4 \quad\text{and}\quad y = -\tfrac23 x - 4. \]

Marking scheme — Scale 15D (0, 4, 7, 10, 15)

Consider the solution as consisting of four steps:

Step 1: Uses the area formula to obtain an equation in \(a\) and \(b\), e.g. \(\tfrac12 ab = 12\) or \(ab = 24\).
Step 2: Uses the slope \(-\tfrac23\) to obtain a second equation in \(a\) and \(b\), e.g. \(-b/a = -2/3\) or \(3b = 2a\).
Step 3: Solves for suitable integer values of \(a\) and \(b\).
Step 4: Finds the equations of both lines.

Full credit (15): all four steps correct — both line equations found (e.g. \(2x+3y=12\) and \(2x+3y=-12\), or equivalent forms).

High partial (10): three of the four steps correct, or the equation of one line only.

Mid partial (7): two steps correct.

Low partial (4): work of merit, for example a relevant diagram, or one correct substitution in an area formula or equation-of-a-line formula, or plots \( (0,b) \) with \(b\) a non-zero even integer, or plots \( (a,0) \) with \(a\) a non-zero multiple of 3.

LC Higher Level — Coordinate Geometry of the Line

  • Distance, midpoint and slope: \[ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2},\quad M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right),\quad m=\frac{y_2-y_1}{x_2-x_1}. \]
  • Equation forms: \[ y=mx+c,\qquad y-y_1=m(x-x_1),\qquad ax+by+c=0. \]
  • Parallel and perpendicular lines:
    • Parallel: \(m_1 = m_2\)
    • Perpendicular: \(m_1m_2=-1\)
  • Intersection of lines; slope as rate of change; intercept as initial value.
  • Area of a triangle (two-point version): \[ A=\tfrac12\,\bigl|\,x_1y_2-x_2y_1\,\bigr|. \]
  • Perpendicular distance from \((x_1,y_1)\) to \(ax+by+c=0\): \[ d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}. \]
  • Angle between two lines of slopes \(m_1,m_2\): \[ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|. \]
  • Internal division of the line from \((x_1,y_1)\) to \((x_2,y_2)\) in the ratio \(a:b\): \[ \left(\frac{bx_1+ax_2}{a+b},\ \frac{by_1+ay_2}{a+b}\right). \]
  • Triangle concurriencies:
    • Centroid: \[ G=\left(\tfrac{x_1+x_2+x_3}{3},\,\tfrac{y_1+y_2+y_3}{3}\right) \]
    • Circumcentre: intersection of perpendicular bisectors
    • Orthocentre: intersection of altitudes