Coordinate Geometry • Topic 2 of 3

Section Formula — Internal Division

What is the section formula? The section formula gives the coordinates of the point that divides a line segment joining A($x_1$, $y_1$) and B($x_2$, $y_2$) in a given ratio $m_1:m_2$ (read as $m_1$ to $m_2$). If P divides AB internally so that AP : PB $=m_1:m_2$, then

$P=\left(\dfrac{m_1 x_2+m_2 x_1}{m_1+m_2},\ \dfrac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)$.

Reading the formula correctly. The first part of the ratio, $m_1$, multiplies the far point B; the second part, $m_2$, multiplies the near point A. A reliable memory aid is "cross-multiply": $m_1$ goes with $x_2$, and $m_2$ goes with $x_1$. The denominator is always the sum $m_1+m_2$. Get the pairing wrong and you will land on the reflection of the correct point, so always write the ratio as AP : PB before substituting.

The midpoint formula. When the ratio is $1:1$ the point is exactly halfway along AB. Putting $m_1=m_2=1$ gives

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

which is simply the average of the two endpoints' coordinates. The midpoint is used constantly: to find the centre of a circle given a diameter, to find a missing endpoint, and to test whether a quadrilateral is a parallelogram (its diagonals bisect each other, i.e. share the same midpoint).

Points of trisection. The two points that divide AB into three equal parts are found with ratios $1:2$ (nearer A) and $2:1$ (nearer B). Apply the section formula twice.

Finding the ratio. If you are asked in what ratio P divides AB, let the ratio be $k:1$. Then $x=\dfrac{k\,x_2+x_1}{k+1}$. Solve this single equation for $k$. A negative value of $k$ signals external division (outside the segment), which is beyond the Grade-10 internal formula. The same trick finds where a line or an axis cuts a segment: for the x-axis the dividing point has $y=0$, for the y-axis it has $x=0$ — set up the $y$ (or $x$) equation equal to 0 and solve for $k$.

Where the formula comes from. Drop perpendiculars from A, P and B to the x-axis, meeting it at A$'$, P$'$ and B$'$. Because AP : PB $=m_1:m_2$, similar triangles give A$'$P$'$ : P$'$B$'$ $=m_1:m_2$ on the axis as well. Writing the foot of P at $x$, we get $\dfrac{x-x_1}{x_2-x}=\dfrac{m_1}{m_2}$; cross-multiplying, $m_2(x-x_1)=m_1(x_2-x)$, which rearranges to $x(m_1+m_2)=m_1 x_2+m_2 x_1$, i.e. $x=\dfrac{m_1 x_2+m_2 x_1}{m_1+m_2}$. The same argument on the y-axis gives the $y$-coordinate. So the formula is just similar triangles dressed up in coordinates — it is worth deriving once so the pairing of $m_1$ with $x_2$ stops being something to memorise blindly.

Centroid of a triangle. The medians of a triangle meet at the centroid G, which divides each median in the ratio $2:1$ from the vertex. To see why the averaging formula falls out, take the midpoint D of BC, which is $\left(\dfrac{x_2+x_3}{2},\dfrac{y_2+y_3}{2}\right)$, then divide AD in the ratio $2:1$ from A. The result simplifies to the simple average of all three vertices:

$G=\left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3}\right)$.

If you are given the centroid and two vertices, you can recover the third by undoing the average — multiply the centroid coordinates by 3 and subtract the two known vertices.

Common mistakes to avoid.

  • Swapping the roles of $m_1$ and $m_2$ — remember $m_1$ multiplies the second point $x_2$.
  • Dividing by $m_1-m_2$ instead of $m_1+m_2$ (the minus sign belongs to external division only).
  • Confusing the centroid (divide by 3) with the midpoint (divide by 2).
  • Forgetting to verify with the $y$-coordinate after finding a ratio from the $x$-coordinate.
Point P dividing segment AB internally in the ratio m to nInternal division: AP : PB = m₁ : m₂ABP(x₁, y₁)(x₂, y₂)m₁ (= 1)m₂ (= 2)P = ( (m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂) )Midpoint as the average of the two endpointsMidpoint (ratio 1:1): average the coordinatesxyOA(2,3)B(8,7)M(5,5)M = ( (2+8)/2, (3+7)/2 )M = (5, 5)
Solved Examples
1
Worked Example
Find the coordinates of the point P that divides the segment joining A(4, −1) and B(−2, −3) in the ratio 1 : 2 internally.
Solution
  1. Step 1: $x_1=4,\ y_1=-1,\ x_2=-2,\ y_2=-3,\ m_1=1,\ m_2=2$.
  2. Step 2: $x=\dfrac{m_1 x_2+m_2 x_1}{m_1+m_2}=\dfrac{1(-2)+2(4)}{1+2}=\dfrac{-2+8}{3}=2$.
  3. Step 3: $y=\dfrac{m_1 y_2+m_2 y_1}{m_1+m_2}=\dfrac{1(-3)+2(-1)}{3}=\dfrac{-5}{3}$.

Answer: P$\left(2,\ -\dfrac53\right)$

2
Worked Example
Find the coordinates of the midpoint of the segment joining P(3, 5) and Q(9, 13).
Solution
  1. Step 1: Midpoint $=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$.
  2. Step 2: $x=\dfrac{3+9}{2}=6$.
  3. Step 3: $y=\dfrac{5+13}{2}=9$.

Answer: M(6, 9)

3
Worked Example
Find the coordinates of the points of trisection of the segment joining A(4, −1) and B(−2, −3).
Solution
  1. Step 1: Trisection point P (nearer A) divides AB in $1:2$.
  2. Step 2: $P=\left(\dfrac{1(-2)+2(4)}{3},\dfrac{1(-3)+2(-1)}{3}\right)=\left(2,-\dfrac53\right)$.
  3. Step 3: Trisection point Q (nearer B) divides AB in $2:1$.
  4. Step 4: $Q=\left(\dfrac{2(-2)+1(4)}{3},\dfrac{2(-3)+1(-1)}{3}\right)=\left(0,-\dfrac73\right)$.

Answer: P$\left(2,-\dfrac53\right)$ and Q$\left(0,-\dfrac73\right)$

4
Worked Example
In what ratio does the point P(−4, 6) divide the segment joining A(−6, 10) and B(3, −8)?
Solution
  1. Step 1: Let the ratio be $k:1$. Then $x=\dfrac{k(3)+1(-6)}{k+1}=-4$.
  2. Step 2: $3k-6=-4(k+1)\Rightarrow 3k-6=-4k-4$.
  3. Step 3: $7k=2\Rightarrow k=\dfrac27$, so the ratio is $2:7$.
  4. Step 4: Check with $y$: $\dfrac{2(-8)+7(10)}{2+7}=\dfrac{-16+70}{9}=\dfrac{54}{9}=6$ ✓.

Answer: P divides AB in the ratio 2 : 7

5
Worked Example
In what ratio does the x-axis divide the segment joining A(1, −5) and B(−4, 5)? Also find the point of division.
Solution
  1. Step 1: On the x-axis $y=0$. Let the ratio be $k:1$.
  2. Step 2: $y=\dfrac{k(5)+1(-5)}{k+1}=0\Rightarrow 5k-5=0\Rightarrow k=1$.
  3. Step 3: So the x-axis divides AB in $1:1$ (the midpoint).
  4. Step 4: $x=\dfrac{1(-4)+1(1)}{2}=\dfrac{-3}{2}$, $y=0$.

Answer: Ratio 1 : 1; point of division $\left(-\dfrac32,\ 0\right)$

6
Worked Example
Find the ratio in which the y-axis divides the segment joining A(5, −6) and B(−1, −4). Also find the point.
Solution
  1. Step 1: On the y-axis $x=0$. Let the ratio be $k:1$.
  2. Step 2: $x=\dfrac{k(-1)+1(5)}{k+1}=0\Rightarrow -k+5=0\Rightarrow k=5$.
  3. Step 3: Ratio is $5:1$.
  4. Step 4: $y=\dfrac{5(-4)+1(-6)}{5+1}=\dfrac{-20-6}{6}=\dfrac{-26}{6}=-\dfrac{13}{3}$.

Answer: Ratio 5 : 1; point $\left(0,\ -\dfrac{13}{3}\right)$

7
Worked Example
The centre of a circle is C(2, −3) and one end of a diameter is A(5, 4). Find the other end B.
Solution
  1. Step 1: The centre is the midpoint of the diameter, so $C=\left(\dfrac{5+x}{2},\dfrac{4+y}{2}\right)$.
  2. Step 2: $\dfrac{5+x}{2}=2\Rightarrow 5+x=4\Rightarrow x=-1$.
  3. Step 3: $\dfrac{4+y}{2}=-3\Rightarrow 4+y=-6\Rightarrow y=-10$.

Answer: B(−1, −10)

8
Worked Example
Find the centroid of the triangle whose vertices are A(−1, −3), B(4, −2) and C(−6, 8).
Solution
  1. Step 1: $G=\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)$.
  2. Step 2: $x=\dfrac{-1+4-6}{3}=\dfrac{-3}{3}=-1$.
  3. Step 3: $y=\dfrac{-3-2+8}{3}=\dfrac{3}{3}=1$.

Answer: Centroid G(−1, 1)

9
Worked Example
Two vertices of a triangle are A(1, 2) and B(3, 5), and its centroid is G(3, 3). Find the third vertex C.
Solution
  1. Step 1: Let C be $(x, y)$. Then $\dfrac{1+3+x}{3}=3$ and $\dfrac{2+5+y}{3}=3$.
  2. Step 2: $4+x=9\Rightarrow x=5$.
  3. Step 3: $7+y=9\Rightarrow y=2$.

Answer: C(5, 2)

10
Worked Example
Find the coordinates of the point which divides the segment joining A(−1, 7) and B(4, −3) in the ratio 2 : 3.
Solution
  1. Step 1: $m_1=2,\ m_2=3$, A$(-1,7)$, B$(4,-3)$.
  2. Step 2: $x=\dfrac{2(4)+3(-1)}{2+3}=\dfrac{8-3}{5}=1$.
  3. Step 3: $y=\dfrac{2(-3)+3(7)}{5}=\dfrac{-6+21}{5}=3$.

Answer: Point (1, 3)

11
Worked Example
If the points A(6, 1), B(8, 2), C(9, 4) and D($x$, $y$) are the vertices of a parallelogram taken in order, find D using the property that diagonals bisect each other.
Solution
  1. Step 1: In parallelogram ABCD the diagonals AC and BD share the same midpoint.
  2. Step 2: Midpoint of AC $=\left(\dfrac{6+9}{2},\dfrac{1+4}{2}\right)=\left(\dfrac{15}{2},\dfrac52\right)$.
  3. Step 3: Midpoint of BD $=\left(\dfrac{8+x}{2},\dfrac{2+y}{2}\right)$.
  4. Step 4: Equate $x$: $\dfrac{8+x}{2}=\dfrac{15}{2}\Rightarrow x=7$.
  5. Step 5: Equate $y$: $\dfrac{2+y}{2}=\dfrac52\Rightarrow y=3$.

Answer: D(7, 3)

12
Worked Example
In what ratio is the segment joining A(−2, −3) and B(3, 7) divided by the line $2x+y-4=0$?
Solution
  1. Step 1: Let the line cut AB at P in the ratio $k:1$. Then $P=\left(\dfrac{3k-2}{k+1},\dfrac{7k-3}{k+1}\right)$.
  2. Step 2: P lies on the line, so $2\!\left(\dfrac{3k-2}{k+1}\right)+\dfrac{7k-3}{k+1}-4=0$.
  3. Step 3: Multiply by $(k+1)$: $2(3k-2)+(7k-3)-4(k+1)=0$.
  4. Step 4: $6k-4+7k-3-4k-4=0\Rightarrow 9k-11=0\Rightarrow k=\dfrac{11}{9}$.
  5. Step 5: The ratio is $11:9$.

Answer: The line divides AB in the ratio 11 : 9

Key Points

  • Section formula (internal, ratio $m_1:m_2$): $P=\left(\dfrac{m_1 x_2+m_2 x_1}{m_1+m_2},\dfrac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)$
  • $m_1$ multiplies the far point $x_2$; $m_2$ multiplies the near point $x_1$
  • Midpoint (ratio 1:1): $M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$
  • Points of trisection use ratios $1:2$ and $2:1$
  • To find an unknown ratio, set it as $k:1$ and solve; verify with the other coordinate
  • The x-axis cuts a segment where $y=0$; the y-axis where $x=0$
  • Centroid: $G=\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3}\right)$ — divide by 3, not 2
  • In a parallelogram the diagonals bisect each other (equal midpoints)
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The internal section formula for the $x$-coordinate (ratio $m:n$) is:
Explanation: Standard formula.
Q2.The midpoint of $(x_1,y_1)$ and $(x_2,y_2)$ is:
Explanation: Average of the coordinates.
Q3.The midpoint of $(2,4)$ and $(6,8)$ is:
Explanation: Average each coordinate.
Q4.The midpoint divides a segment in the ratio:
Explanation: Equal halves.
Practice more MCQs → 📝 Assignment · 35 marks