MHT-CET Maths · Straight Line

Section Formula, Midpoints and Rectangles

The point dividing AB in m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)); the midpoint is the 1 : 1 case, and it is the centre of any rectangle's diagonal and the circumcentre of a right triangle.

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Why this matters

11 PYQs, one HARD — the smallest page in the chapter and one of the cheapest. A ratio on a segment, the centre of a rectangle given by its side lines, the two missing vertices of a rectangle from one diagonal and the line they lie on (set in 2024 and 2025), a circumcentre, and the ratio in which the origin divides a segment between two parallel lines. Each is one formula and one substitution.

Concept 1 of 2: Section Formula: Internal and External Division

Dividing

AB

internally in

m : n

weights

B

m

and

A

n

. External division flips one sign. When a line through the origin meets two PARALLEL lines, the origin divides the segment in the ratio of the perpendicular distances from the origin to the two lines.

Definition

$(1, 1)$ , $(2, 4)$ in $3 : 2$ : $\left(\dfrac{6 + 2}{5}, \dfrac{12 + 2}{5}\right) = \left(\dfrac85, \dfrac{14}{5}\right)$ . On $2x + y = k$ : $k = 6$ , so $(k + 1) : (k - 1) = 7 : 5$ .
$4x + 3y = 10$ and $8x + 6y + 5 = 0$ are parallel, at distances $2$ and $\tfrac12$ from the origin, on opposite sides; a line through $O$ meets them at $A$ , $B$ with $AO : OB = 2 : \tfrac12 = 4 : 1$ .
External division in $m : n$ : $\left(\dfrac{mx_2 - nx_1}{m - n}, \dfrac{my_2 - ny_1}{m - n}\right)$ .
$P$ divides $A(a, 0)$ , $B(0, b)$ in $1 : 2$ at $(-4, 1)$ : $\dfrac{2a}{3} = -4$ , $\dfrac{b}{3} = 1$ — the intercepts follow.

Section formula

P = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}\right) \quad(\text{internal});\qquad \frac{AO}{OB} = \frac{d_1}{d_2} \text{ between parallel lines}

Diagram · section formula (internal vs external), m : n = 2 : 1

Internal: P = (m·b + n·a)/(m + n) sits between A and B. External: Q = (m·b − n·a)/(m − n) sits beyond B — the minus sign is what pushes it outside. The midpoint is the m = n case, (a + b)/2.

Worked example

Find the point dividing

(2, -1)

and

(7, 9)

internally in

2 : 3

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift I · Q146Easy

Example 1 · Straight Line · Section Formula, Midpoints and Rectangles

If the line

2x+y= k

passes though the point which divides the line segment joining the points

(1,1)

and

(2,4)

internally in the ratio

3:2

then,

(k+ 1):(k- 1) =

Swapping the weights

m : n

the weight

m

goes on the SECOND point.

\left(\dfrac{3 \cdot 1 + 2 \cdot 2}{5}, \dots\right)

gives

\left(\dfrac75, \dfrac{11}{5}\right)

and a different

k

Drill 1 more on section formula: internal and external division

Concept 2 of 2: Midpoints Everywhere: Rectangle Centres, Missing Vertices and the Circumcentre

A rectangle's diagonals bisect each other, so the centre is the midpoint of either diagonal; a missing vertex lies on the given line AND makes a right angle with the known diagonal's ends. The circumcentre is equidistant from the vertices — solve two perpendicular-bisector equations.

Definition

Sides $x = 8, x = 10, y = 11, y = 12$ : the diagonals meet at the centre $\left(9, \tfrac{23}{2}\right)$ .
Opposite vertices $(1, 3)$ , $(5, 1)$ , other two on $y = 2x + c$ : the centre $(3, 2)$ lies on that line, so $c = -4$ . A vertex $(x, 2x - 4)$ sees the diagonal at a right angle: slope to $(1,3)$ times slope to $(5,1)$ is $-1$ , giving $x^2 - 6x + 8 = 0$ , $x = 2, 4$ : $(2, 0)$ and $(4, 4)$ .
Circumcentre of $(-2, 3)$ , $(6, -1)$ , $(4, 3)$ : equate distances, $2h - k = 3$ and $h - 2k = -3$ , so $(1, -1)$ . For a right triangle it is the midpoint of the hypotenuse.
The centre trick — 'the other diagonal passes through the midpoint of the first' — is what fixes $c$ without knowing either missing vertex.

Rectangle centre

\text{centre} = \text{midpoint of a diagonal};\qquad \text{right angle at a vertex: } m_{VA}\, m_{VC} = -1

Worked example

(0, 0)

and

(6, 2)

are opposite vertices of a rectangle whose other vertices lie on

y = x + c

. Find

c

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q148Moderate

Example 2 · Straight Line · Section Formula, Midpoints and Rectangles

The points

(1,3)

and

(5,1)

are two opposite vertices of a rectangle. The other two vertices are lie on the line

y= 2x+ c

where c is the constant, then co-ordinates of other two vertices are

Solving for the vertices without fixing c first

The line

y = 2x + c

has an unknown; the midpoint of the known diagonal lies on it and gives

c = -4

in one line. Without that step the right-angle condition has two unknowns.

Drill 3 more on midpoints everywhere: rectangle centres, missing vertices and the circumcentre

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Section Formula: Internal and External Division
Section formula
$P = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}\right) \quad(\text{internal});\qquad \frac{AO}{OB} = \frac{d_1}{d_2} \text{ between parallel lines}$
Midpoints Everywhere: Rectangle Centres, Missing Vertices and the Circumcentre
Rectangle centre
$\text{centre} = \text{midpoint of a diagonal};\qquad \text{right angle at a vertex: } m_{VA}\, m_{VC} = -1$

Watch out for (2)

Swapping the weights
→ Section Formula: Internal and External Division
Solving for the vertices without fixing c first
→ Midpoints Everywhere: Rectangle Centres, Missing Vertices and the Circumcentre

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