Section and Mid-Point Formulae notes: ICSE Class 10 Mathematics

1. Dividing a segment internally

A point $P$ divides $AB$ internally in the ratio $m_1 : m_2$ when $P$ lies on the segment $AB$ and $AP : PB = m_1 : m_2$ .
$m_1$ belongs to the part next to $A$ and $m_2$ to the part next to $B$ ; if $m_1 < m_2$ , $P$ is nearer to $A$ .
A fraction $\dfrac{a}{b}$ of the way from $A$ to $B$ means $AP : PB = a : (b - a)$ : a quarter of the way is $1 : 3$ .
(Division of a segment externally, with $P$ outside $AB$ , is not in the ICSE 2027 syllabus.)

Example: $A(1, 1)$ , $B(10, 7)$ : the point one-third of the way along divides $AB$ in the ratio $1 : 2$ ; it is $(4, 3)$ , since $x = \dfrac{1 \times 10 + 2 \times 1}{1 + 2} = \dfrac{12}{3} = 4$ and $y = \dfrac{1 \times 7 + 2 \times 1}{1 + 2} = \dfrac{9}{3} = 3$ .

segment AB in three equal parts

$AP$ is $1$ part, $PB$ is $2$ parts: $P$ divides $AB$ internally in the ratio $1 : 2$.

segment AB

$P$ splits $AB$ into $2$ parts and $3$ parts: $AP : PB = 2 : 3$.

Key terms

Internal division: A point $P$ on the segment $AB$ divides it internally in the ratio $AP : PB$ .
Ratio $m_1 : m_2$: The comparison of the two parts $AP$ and $PB$ ; only its proportion matters, so $2 : 2$ is the same as $1 : 1$ .

2. The section formula

The point dividing the join of $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m_1 : m_2$ is

P\left(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \ \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)

Why: the horizontal and vertical steps from $A$ to $P$ and from $P$ to $B$ form similar right triangles, so $\dfrac{x - x_1}{x_2 - x} = \dfrac{m_1}{m_2}$ ; solve for $x$ (and the same for $y$ ).
Cross pairing: $m_1$ multiplies the far end $B$ , $m_2$ multiplies $A$ ; then divide by $m_1 + m_2$ .

Example: $A(-5, 9)$ , $B(5, -6)$ , ratio $2 : 3$ : $x = \dfrac{2 \times 5 + 3 \times (-5)}{2 + 3} = \dfrac{-5}{5} = -1$ and $y = \dfrac{2 \times (-6) + 3 \times 9}{2 + 3} = \dfrac{15}{5} = 3$ , so $P$ is $(-1, 3)$ .

segment AB
triangle AKP
triangle PLB

Similar triangles: $\dfrac{AK}{PL} = \dfrac{KP}{LB} = \dfrac{AP}{PB}$.

The same four steps give the $x$-coordinate and the $y$-coordinate.

segment AB

$AP : PB = 2 : 3$ gives $P(-1, 3)$.

Formulas

Section formula, x-coordinate
$x = \dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}$
Section formula, y-coordinate
$y = \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}$

Key terms

Section formula: The formula for the point dividing a segment internally in a given ratio.

3. Points of trisection

The two points of trisection of $AB$ cut it into three equal parts: the one nearer $A$ divides $AB$ in the ratio $1 : 2$ , the one nearer $B$ in the ratio $2 : 1$ .
The trisection point nearer $B$ is the mid-point of the other trisection point and $B$ — a quick check.

Example: for $A(-7, 2)$ and $B(8, -4)$ : ratio $1 : 2$ gives $x = \dfrac{1 \times 8 + 2 \times (-7)}{1 + 2} = \dfrac{-6}{3} = -2$ , $y = \dfrac{1 \times (-4) + 2 \times 2}{1 + 2} = \dfrac{0}{3} = 0$ , so $(-2, 0)$ ; ratio $2 : 1$ gives $x = \dfrac{2 \times 8 + 1 \times (-7)}{2 + 1} = \dfrac{9}{3} = 3$ , $y = \dfrac{2 \times (-4) + 1 \times 2}{2 + 1} = \dfrac{-6}{3} = -2$ , so $(3, -2)$ .

segment AB

$P$: ratio $1 : 2$; $Q$: ratio $2 : 1$.

A fraction $\dfrac{a}{b}$ of the way from $A$ means $AP : PB = a : (b - a)$.
Where $P$ is on $AB$	Ratio $AP : PB$
halfway (the mid-point)	$1 : 1$
one-third of the way from $A$ (trisection point nearer $A$ )	$1 : 2$
two-thirds of the way from $A$ (trisection point nearer $B$ )	$2 : 1$
one-quarter of the way from $A$	$1 : 3$
two-fifths of the way from $A$	$2 : 3$

Key terms

Points of trisection: The two points that divide a segment into three equal parts.

4. Finding the ratio: a point, an axis or a line

Unknown ratio: call it $k : 1$ , so the point is $\left(\dfrac{kx_2 + x_1}{k + 1}, \dfrac{ky_2 + y_1}{k + 1}\right)$ . One coordinate gives $k$ ; the other is a check. Write $k : 1$ in whole numbers ( $k = \dfrac{2}{3}$ means $2 : 3$ ).
An axis: on the $x$ -axis $y = 0$ ; on the $y$ -axis $x = 0$ .
A line $ax + by = c$ : substitute the $k : 1$ point into the equation and solve for $k$ .
A positive $k$ means the division is internal (the ends are on opposite sides of the axis or line).

Example: $P(1, -1)$ on the join of $A(-5, 3)$ and $B(4, -3)$ : $\dfrac{4k + (-5)}{k + 1} = 1 \Rightarrow 4k - 5 = k + 1 \Rightarrow 3k = 6 \Rightarrow k = 2$ , so $AP : PB = 2 : 1$ . The line $x + y = 3$ cuts the join of $(-2, 1)$ and $(7, 4)$ where $(7k - 2) + (4k + 1) = 3(k + 1) \Rightarrow 11k - 1 = 3k + 3 \Rightarrow 8k = 4 \Rightarrow k = \dfrac{1}{2}$ , in the ratio $1 : 2$ .

segment AB

On the $y$-axis $x = 0$; on the $x$-axis $y = 0$.

One unknown, $k$, needs only one equation.

segment AB

$AP : PB = 2 : 1$.

segment AB

The $y$-axis cuts $AB$ in the ratio $1 : 3$, the $x$-axis in the ratio $5 : 7$.

segment AB
line x + y = 3

The line $x + y = 3$ cuts $AB$ in the ratio $1 : 2$ at $P(1, 2)$.

Every ratio problem becomes one equation in $k$.
You are given	Do this
the ends and the ratio	section formula for $x$ and for $y$
the ends and a point on $AB$	ratio $k : 1$ from one coordinate, check with the other
an axis that crosses $AB$	$x = 0$ (the $y$ -axis) or $y = 0$ (the $x$ -axis), then solve for $k$
a line $ax + by = c$ that crosses $AB$	put the $k : 1$ point into the equation of the line
one coordinate of a point on $AB$	find $k$ from it, then the other coordinate

Formulas

Point in the ratio k : 1
$x = \dfrac{kx_2 + x_1}{k + 1}$

Key terms

Ratio $k : 1$: Writing an unknown ratio with second part $1$ leaves a single unknown, $k$ .

5. The mid-point formula and a missing end

The mid-point is the $1 : 1$ section point, the average of the ends:

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

A missing end: if $M(x, y)$ is the mid-point and $A(x_1, y_1)$ one end, the other end is $B(2x - x_1, 2y - y_1)$ .

Example: the mid-point of $(-7, 4)$ and $(3, -10)$ is $(-2, -3)$ , since $\dfrac{-7 + 3}{2} = \dfrac{-4}{2} = -2$ and $\dfrac{4 + (-10)}{2} = \dfrac{-6}{2} = -3$ . If $A$ is $(6, -1)$ and $M$ is $(2, 3)$ , then $2 \times 2 - 6 = -2$ and $2 \times 3 - (-1) = 7$ , so $B$ is $(-2, 7)$ .

segment AB

$M$ is the average of the ends: $\left(\dfrac{-4 + 6}{2}, \dfrac{-1 + 5}{2}\right) = (1, 2)$.

A to M
M to B, the same step again

From $A$ to $M$, then the same step again: $B = (2x - x_1, 2y - y_1)$.

segment AB

$B(-2, 7)$: check that $M$ is halfway.

Formulas

Mid-point formula
$M\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$
Other end from the mid-point
$B(2x - x_1, 2y - y_1)$

Key terms

Mid-point: The point of a segment equidistant from its ends; it divides the segment in the ratio $1 : 1$ .

6. Parallelograms

The diagonals of a parallelogram bisect each other, so they have the same mid-point.
For a parallelogram $ABCD$ (vertices in order), the mid-point of $AC$ is the mid-point of $BD$ ; so $D$ has coordinates $A + C - B$ .
Showing that the diagonals share their mid-point is also a way to prove that a quadrilateral is a parallelogram.

Example: $A(-2, -1)$ , $B(4, 0)$ , $C(5, 4)$ : the mid-point of $AC$ is $\left(\dfrac{3}{2}, \dfrac{3}{2}\right)$ , and then $2 \times \dfrac{3}{2} - 4 = -1$ and $2 \times \dfrac{3}{2} - 0 = 3$ , so $D$ is $(-1, 3)$ .

parallelogram PQRS
diagonal PR
diagonal QS

The diagonals of a parallelogram bisect each other.

parallelogram ABCD
diagonal AC
diagonal BD

$M$ is the mid-point of $AC$ and of $BD$.

Key terms

Bisect: Cut into two equal parts; the diagonals of a parallelogram bisect each other.

7. The centroid of a triangle

A median joins a vertex to the mid-point of the opposite side. The three medians meet at the centroid $G$ , which divides each median in the ratio $2 : 1$ from the vertex.
Formula: the section formula with ratio $2 : 1$ on a median gives the average of the vertices:

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

Missing vertex: $x_3 = 3x - x_1 - x_2$ and $y_3 = 3y - y_1 - y_2$ .
Vertices from the mid-points $D$ , $E$ , $F$ of $BC$ , $CA$ , $AB$ : $AFDE$ is a parallelogram, so $A$ is $E + F - D$ (and similarly for $B$ and $C$ ). A triangle and its triangle of mid-points have the same centroid.

Example: for $(-3, 5)$ , $(7, 1)$ , $(2, -9)$ : $\dfrac{-3 + 7 + 2}{3} = \dfrac{6}{3} = 2$ and $\dfrac{5 + 1 + (-9)}{3} = \dfrac{-3}{3} = -1$ , so $G$ is $(2, -1)$ . With $G(1, 2)$ , $A(-4, 3)$ , $B(5, -1)$ : $3 \times 1 - (-4) - 5 = 2$ and $3 \times 2 - 3 - (-1) = 4$ , so $C$ is $(2, 4)$ .

triangle ABC
median AD

$AG : GD = 2 : 1$.

Twice the mid-point of $BC$ is the sum of $B$ and $C$, so the result is an average.

triangle ABC
median AD

$G(2, -1)$ by the formula and by the median agree.

triangle ABC
median from C

$C = (3 \times 1 - (-4) - 5, \ 3 \times 2 - 3 - (-1)) = (2, 4)$.

triangle ABC
triangle DEF of mid-points

Triangles $ABC$ and $DEF$ share the centroid $(1, 1)$.

triangle ABC
parallelogram AFDE

$AFDE$ is a parallelogram, so $A$ is its fourth vertex.

Five situations, five one-line formulas.
You know	You want	Use
both ends	the mid-point	$\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$
one end and the mid-point	the other end	$(2x - x_1, 2y - y_1)$
three vertices of a parallelogram	the fourth	diagonals share their mid-point
three vertices	the centroid	$\left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)$
two vertices and the centroid	the third vertex	$(3x - x_1 - x_2, 3y - y_1 - y_2)$

Formulas

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

Key terms

Median: The segment from a vertex of a triangle to the mid-point of the opposite side.
Centroid: The point where the three medians meet; it divides each median in the ratio $2 : 1$ from the vertex.

Exam tips

How marks are awarded

Section point (2 marks): correct substitution 1; both coordinates 1.
Ratio from a point, an axis or a line (2–3 marks): equation in $k$ 1; ratio in whole numbers 1; point of division 1.
Mid-point, missing end, centroid, missing vertex (1–2 marks): formula with substitution 1; answer 1.
10-mark question: parts (a), (b), (c), usually 3 + 3 + 4, often mixing the section formula, mid-points and the centroid; show all working.

Common mistakes

Pairing $m_1$ with $A$ instead of $B$ (that gives the point in the ratio $m_2 : m_1$ ).
Reading "a quarter of the way from $A$ " as $1 : 4$ instead of $1 : 3$ .
Setting $y = 0$ for the $y$ -axis (on the $y$ -axis it is $x$ that is $0$ ).
Leaving a ratio as a fraction: $k = \dfrac{3}{5}$ means $3 : 5$ .
Halving instead of doubling when finding the other end from the mid-point.
Dividing by $2$ instead of $3$ for the centroid, or giving the median ratio as $1 : 2$ from the vertex.

1. Dividing a segment internally

Key terms

2. The section formula

Formulas

Key terms

3. Points of trisection

Key terms

4. Finding the ratio: a point, an axis or a line

Formulas

Key terms

5. The mid-point formula and a missing end

Formulas

Key terms

6. Parallelograms

Key terms

7. The centroid of a triangle

Formulas

Key terms

Exam tips

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