MHT-CET Maths · Vectors

Vector Geometry — Section Formula, Triangle, and Parallelogram

Using the section formula (internal and external), the centroid and median identities, the triangle-centre formulas (incentre, orthocentre), and parallelogram relations to locate points and ratios from position vectors.

All Vectors notes MHT-CET Maths strategy

Why this matters

This is the bread-and-butter geometry strand of MHT-CET Vectors — about a dozen PYQs across the years, roughly half of them HARD. The two engines are the section formula (a point dividing a segment in a given ratio) and the centroid as the average of vertex position vectors; from those, medians, cevian-intersection ratios, the incentre, and parallelogram classification all follow. Get the internal-vs-external sign right and decide which point is weighted m versus n, and most of these resolve to a few lines of vector algebra.

Concept 1 of 5

Section formula — internal, external, and midpoint

Intuition

A point that splits the segment joining

A

and

B

in the ratio

m:n

sits at a weighted average of the two endpoint position vectors. For internal division the weights add; for external division they subtract. The midpoint is the symmetric case

m=n

Definition

Let $A$ and $B$ have position vectors $\vec{a}$ and $\vec{b}$ , and let $R$ divide $AB$ in the ratio $m:n$ .

Internal division ( $R$ between $A$ and $B$ ): $\vec{r} = \dfrac{m\vec{b} + n\vec{a}}{m + n}$ .
External division ( $R$ on the line, outside $AB$ ): $\vec{r} = \dfrac{m\vec{b} - n\vec{a}}{m - n}$ .
Midpoint ( $m = n = 1$ ): $\vec{r} = \dfrac{\vec{a} + \vec{b}}{2}$ .

In the internal form, the far endpoint $\vec{b}$ carries the weight $m$ and the near endpoint $\vec{a}$ carries $n$ , where $AR : RB = m : n$ . Memorise the form rather than a side-story.

Section formula (internal / external / midpoint)

\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n} \qquad \vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n} \qquad \vec{r} = \frac{\vec{a} + \vec{b}}{2}

$\vec{a}, \vec{b}$ position vectors of the endpoints $A, B$
$m : n$ ratio $AR : RB$ in which $R$ divides the segment
$\vec{r}$ position vector of the dividing point $R$

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 position vector of the point

R

that divides the segment joining

A(\vec{a} = 2\hat{i} + \hat{j})

and

B(\vec{b} = 5\hat{i} + 7\hat{j})

internally in the ratio

2:1

Internal division: $\vec{r} = \dfrac{m\vec{b} + n\vec{a}}{m + n}$ with $m = 2$ , $n = 1$ .
$\vec{r} = \dfrac{2(5\hat{i} + 7\hat{j}) + 1(2\hat{i} + \hat{j})}{3} = \dfrac{12\hat{i} + 15\hat{j}}{3}$ .
$\vec{r} = 4\hat{i} + 5\hat{j}$ .

Answer:

\vec{r} = 4\hat{i} + 5\hat{j}

Practice this concept4 quick reps

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Midpoint of $\vec{a}$ and $\vec{b}$ ?
2.
Internal-division formula for ratio $m:n$ ?
3.
External-division formula for ratio $m:n$ ?
4.
Point dividing $A(0,0)$ , $B(6,3)$ internally $1:2$ ?

Internal adds, external subtracts

The only difference between the two formulas is the sign in the denominator (and numerator): internal uses

m + n

, external uses

m - n

. An external-division question with the internal formula (or vice versa) is the single most common slip — read whether

R

lies between the points or beyond them.

Which point gets the weight $m$ ?

\dfrac{m\vec{b} + n\vec{a}}{m+n}

, the FAR endpoint

\vec{b}

carries

m

and the NEAR endpoint

\vec{a}

carries

n

, where the ratio is

AR:RB = m:n

. Swapping the weights places

R

at the mirror point. When unsure, sanity-check: a

2:1

point should sit closer to

B

Concept 2 of 5

Centroid and median identities

Intuition

The centroid of a triangle is just the average of the three vertex position vectors — the balance point of the three corners. A median runs from a vertex to the midpoint of the opposite side, so its vector is built from the midpoint formula. The centroid lies on every median, cutting each in the ratio

2:1

from the vertex.

Definition

For triangle $ABC$ with vertices $\vec{a}, \vec{b}, \vec{c}$ :

Centroid: $\vec{g} = \dfrac{\vec{a} + \vec{b} + \vec{c}}{3}$ .
**Median through $A$ ** to the midpoint $D$ of $BC$ : $\vec{AD} = \dfrac{\vec{b} + \vec{c}}{2} - \vec{a}$ , equivalently $\vec{AD} = \dfrac{\vec{AB} + \vec{AC}}{2}$ .
The centroid divides each median in ratio $2:1$ (vertex to centroid : centroid to midpoint).
Tetrahedron centroid (four vertices): $\vec{g} = \dfrac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}$ .

Centroid and median vector

\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \qquad \vec{AD} = \frac{\vec{AB} + \vec{AC}}{2} \qquad \vec{g}_{\text{tetra}} = \frac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}

$\vec{a}, \vec{b}, \vec{c}$ position vectors of the vertices
$\vec{g}$ position vector of the centroid
$D$ midpoint of $BC$ ; $\vec{AD}$ is the median from $A$

Diagram · closed loop & centroid

Walking the edges A→B→C→A returns you to the start, so AB + BC + CA = 0. The three medians meet at the centroid G = (a + b + c)/3, the average of the vertices' position vectors.

Worked example

The vectors

\vec{AB} = 2\hat{i} + 6\hat{j} + 3\hat{k}

and

\vec{AC} = 4\hat{i} - 2\hat{j} - \hat{k}

are two sides of triangle

ABC

. Find the length of the median through

A

Median vector: $\vec{AD} = \dfrac{\vec{AB} + \vec{AC}}{2} = \dfrac{6\hat{i} + 4\hat{j} + 2\hat{k}}{2} = 3\hat{i} + 2\hat{j} + \hat{k}$ .
Length: $|\vec{AD}| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{9 + 4 + 1}$ .
$= \sqrt{14}$ .

Answer:

|\vec{AD}| = \sqrt{14}

units

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the centroid of the triangle with vertices

A(1, 2, -1)

B(3, 0, 4)

C(-1, 4, 0)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Centroid formula for $\vec{a}, \vec{b}, \vec{c}$ ?
2.
Median vector through $A$ in terms of the sides?
3.
The centroid divides each median in what ratio (from the vertex)?
4.
Centroid of $A(0,0,0)$ , $B(3,0,0)$ , $C(0,6,0)$ ?

From the bank · past-year question

Example 2VectorsHARD

If D, E and F are the mid-points of the sides BC, CA and AB of triangle ABC respectively, then

\overrightarrow{AD}+\frac{2}{3}\overrightarrow{BE}+\frac{1}{3}\overrightarrow{CF}=

[Q145 · 9th May Shift 2 · 2024]

Median length $\neq$ half the side it bisects

The median through

A

\dfrac{\vec{AB} + \vec{AC}}{2}

, NOT

\dfrac{1}{2}\vec{BC}

. Take half of the SUM of the two adjacent side-vectors, then take its magnitude — don't halve the opposite side's length.

Centroid uses position vectors, not side vectors

\vec{g} = \dfrac{\vec{a} + \vec{b} + \vec{c}}{3}

needs the position vectors of the vertices. If a problem hands you only

\vec{AB}

and

\vec{AC}

, the centroid relative to

A

\dfrac{\vec{AB} + \vec{AC}}{3}

— a different (and frequently tested) form.

Drill 2 more on centroid and median identities

Concept 3 of 5

Finding the ratio, collinearity, and cevian intersection

Intuition

Run the section formula in reverse: when you already know a point on a segment, equating coefficients tells you the ratio in which it divides the segment. The same idea proves collinearity — three points are collinear exactly when one divides the join of the other two in some ratio — and locates the intersection of two cevians by writing the meeting point two ways and matching.

Definition

If $R$ lies on line $AB$ with $\vec{r} = \dfrac{m\vec{b} + n\vec{a}}{m+n}$ , then comparing the two sides recovers $m:n$ . Collinearity: $A, B, C$ are collinear iff $\vec{c} = \lambda\vec{a} + \mu\vec{b}$ with $\lambda + \mu = 1$ (i.e. $C$ divides $AB$ in ratio $\mu:\lambda$ ). Cevian intersection: write the meeting point $G$ as a section point of cevian 1 AND of cevian 2, equate, and solve the resulting linear system for the two parameters.

Ratio recovery and external division

\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n} \;\Rightarrow\; \text{ratio } m:n \qquad \vec{r}_{\text{ext}} = \frac{m\vec{b} - n\vec{a}}{m - n}

$m : n$ the ratio recovered by comparing coefficients
$\vec{r}_{\text{ext}}$ external-division point — used for one branch of perpendicular-cevian problems

Worked example

In what ratio does the point

R = 3\hat{i} + 5\hat{j}

divide the segment joining

A = \hat{i} + \hat{j}

and

B = 5\hat{i} + 9\hat{j}

Let $R$ divide $AB$ internally in ratio $k:1$ : $\vec{r} = \dfrac{k\vec{b} + \vec{a}}{k + 1}$ .
Compare the $\hat{i}$ components: $3 = \dfrac{5k + 1}{k + 1} \Rightarrow 3k + 3 = 5k + 1 \Rightarrow k = 1$ .
Check the $\hat{j}$ components: $\dfrac{9(1) + 1}{2} = 5$ ✓ — consistent, so $k = 1$ .

Answer:

R

divides

AB

in ratio

1:1

(it is the midpoint).

Practice this conceptself-check · 4 quick reps

Try it yourself

Show that

A = \hat{i} + 2\hat{j}

B = 3\hat{i} + 4\hat{j}

C = 4\hat{i} + 5\hat{j}

are collinear, and find the ratio in which

B

divides

AC

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$A, B, C$ collinear with $\vec{c} = \lambda\vec{a} + \mu\vec{b}$ requires?
2.
Recover the ratio: equate the ____ of the two section-formula expressions.
3.
External division of $AB$ in ratio $m:n$ ?
4.
Midpoint divides a segment in ratio?

From the bank · past-year question

Example 3VectorsHARD

In triangle ABC, the point P divides BC internally in the ratio 3 : 4 and Q divides CA internally in the ratio 5 : 3. If AP and BQ intersect in a point G, then G divides AP internally in the ratio

[Shift || · 2025]

Internal and external points use the SAME magnitude of ratio

When

R

and

S

divide

PQ

internally and externally in the same ratio

2:3

, use

\vec{OR} = \dfrac{2\vec{q} + 3\vec{p}}{5}

and

\vec{OS} = \dfrac{2\vec{q} - 3\vec{p}}{-1} = 3\vec{p} - 2\vec{q}

. Forgetting the sign flip in the external denominator is the classic error in OR

\perp

OS problems.

Cevian-intersection ratio is asked along ONE cevian

AP

and

BQ

meet at

G

, the question wants the ratio

AG:GP

(along

AP

) — not

BG:GQ

. Pin which cevian you are reporting the ratio on, and don't invert it (a

5:7

answer reads as

7:5

if you measure from the wrong end).

Drill 1 more on finding the ratio, collinearity, and cevian intersection

Concept 4 of 5

Incentre, orthocentre, and the angle bisector

Intuition

The triangle centres are weighted (or unweighted) averages of the vertices. The incentre weights each vertex by the length of the OPPOSITE side, because it sits where the three internal angle bisectors meet. The orthocentre is where the altitudes meet, characterised by perpendicularity of a vertex-to-point vector against the opposite side. The internal bisector of two vectors points along the sum of their UNIT vectors.

Definition

For triangle $ABC$ with vertices $\vec{a}, \vec{b}, \vec{c}$ and opposite side lengths $a = |BC|$ , $b = |CA|$ , $c = |AB|$ :

Incentre: $\vec{I} = \dfrac{a\vec{a} + b\vec{b} + c\vec{c}}{a + b + c}$ — each vertex weighted by its OPPOSITE side.
Orthocentre: the point $H$ with $(\vec{a} - \vec{h})\cdot(\vec{b} - \vec{c}) = 0$ and $(\vec{b} - \vec{h})\cdot(\vec{c} - \vec{a}) = 0$ (each altitude $\perp$ the opposite side).
Internal angle bisector of $\angle AOB$ : along $\hat{a} + \hat{b} = \dfrac{\vec{a}}{|\vec{a}|} + \dfrac{\vec{b}}{|\vec{b}|}$ — equal coefficients on the two unit vectors.

Incentre and angle bisector

\vec{I} = \frac{a\vec{a} + b\vec{b} + c\vec{c}}{a + b + c} \qquad \text{bisector of }\angle AOB \parallel \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|}

$a, b, c$ lengths of sides opposite $A, B, C$ : $a = |BC|$ , etc.
$\vec{I}$ position vector of the incentre
$\hat{a} + \hat{b}$ sum of unit vectors — the internal-bisector direction

Worked example

Find the incentre of the triangle with vertices

A(0, 0)

B(4, 0)

C(0, 3)

Side lengths (opposite each vertex): $a = |BC| = \sqrt{4^2 + 3^2} = 5$ , $b = |CA| = 3$ , $c = |AB| = 4$ .
$\vec{I} = \dfrac{a\vec{a} + b\vec{b} + c\vec{c}}{a + b + c} = \dfrac{5(0,0) + 3(4,0) + 4(0,3)}{5 + 3 + 4}$ .
$= \dfrac{(12, 12)}{12} = (1, 1)$ .

Answer:Incentre

= (1, 1)

Practice this conceptself-check · 4 quick reps

Try it yourself

The internal bisector of

\angle AOB

, where

\vec{OA} = \vec{a}

and

\vec{OB} = \vec{b}

, is given by

x\dfrac{\vec{a}}{|\vec{a}|} + y\dfrac{\vec{b}}{|\vec{b}|}

. What relation must

x

and

y

satisfy?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Incentre weight on vertex $A$ ?
2.
The orthocentre is the meeting point of the triangle's ____.
3.
Internal bisector of $\angle AOB$ points along?
4.
Centroid vs incentre: which uses side-length weights?

From the bank · past-year question

Example 4VectorsMODERATE

The incentre of the triangle whose vertices are

P(0,3,0)

Q(0,0,4)

and

R(0,3,4)

[Q147 · 2nd May Shift 2 · 2023]

Incentre weights are OPPOSITE side lengths

Vertex

A

is weighted by

a = |BC|

— the side facing

A

, not

|AB|

|AC|

. Mis-pairing the weight with an adjacent side is the standard incentre trap; label the sides

a, b, c

opposite

A, B, C

first.

Bisector uses unit vectors — sum, not difference

The INTERNAL bisector is along

\hat{a} + \hat{b}

(sum of unit vectors); the EXTERNAL bisector is along

\hat{a} - \hat{b}

. Forgetting to normalise (using

\vec{a} + \vec{b}

instead of

\hat{a} + \hat{b}

) gives the wrong direction unless

|\vec{a}| = |\vec{b}|

Orthocentre $\neq$ centroid $\neq$ circumcentre

A perpendicularity condition like

(\vec{a} - \vec{d})\cdot(\vec{b} - \vec{c}) = 0

says

DA \perp BC

— an ALTITUDE, so

D

is the orthocentre. Equal distances to the vertices would mean circumcentre; equal angle-bisector weighting means incentre. Read which condition is given.

Drill 4 more on incentre, orthocentre, and the angle bisector

Concept 5 of 5

Triangle and parallelogram applications

Intuition

Many geometry PYQs are dressed-up dot-product or section-formula problems. A right angle at a vertex is a dot product of the two side-vectors equal to zero. Classifying a quadrilateral from four position vectors is a checklist: equal opposite sides means parallelogram; equal diagonals adds rectangle; perpendicular diagonals adds rhombus.

Definition

Right angle at a vertex: $\angle A = 90^\circ$ iff $\vec{AB} \cdot \vec{AC} = 0$ . Quadrilateral classification from vertices $P, Q, R, S$ (in order):

Opposite sides equal as vectors ( $\vec{PQ} = \vec{SR}$ ) $\Rightarrow$ parallelogram.
A parallelogram with $|\vec{PR}| = |\vec{QS}|$ (equal diagonals) $\Rightarrow$ rectangle.
A parallelogram with $\vec{PR} \cdot \vec{QS} = 0$ (perpendicular diagonals) $\Rightarrow$ rhombus.
Both equal AND perpendicular diagonals $\Rightarrow$ square.

Right-angle test and parallelogram diagonals

\angle A = 90^\circ \iff \vec{AB}\cdot\vec{AC} = 0 \qquad \vec{PR} = \vec{PQ} + \vec{QR}, \;\; \vec{QS} = \vec{QP} + \vec{PS}

$\vec{AB}, \vec{AC}$ the two side-vectors leaving the right-angle vertex
$\vec{PR}, \vec{QS}$ diagonals of quadrilateral $PQRS$

Diagram · parallelogram diagonals = a + b and a − b

From a shared corner, sides a and b span the parallelogram. The diagonal from that corner is a + b; the diagonal between the side tips is a − b. They bisect each other, and |a + b|² + |a − b|² = 2(|a|² + |b|²).

Worked example

Triangle

ABC

has

A = (1, 0, 2)

B = (3, -1, 1)

C = (2, t, 3)

. For what value of

t

is the angle at

A

a right angle?

Side-vectors at $A$ : $\vec{AB} = (2, -1, -1)$ , $\vec{AC} = (1, t, 1)$ .
Right angle at $A$ : $\vec{AB}\cdot\vec{AC} = 0 \Rightarrow (2)(1) + (-1)(t) + (-1)(1) = 0$ .
$2 - t - 1 = 0 \Rightarrow t = 1$ .

Answer:

t = 1

Practice this conceptself-check · 4 quick reps

Try it yourself

Are the points

P(0,0)

Q(4,0)

R(5,3)

S(1,3)

the vertices of a parallelogram (in order)?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Right angle at $A$ iff $\vec{AB}\cdot\vec{AC} = ?$
2.
Equal opposite side-vectors imply which quadrilateral?
3.
Parallelogram with perpendicular diagonals is a?
4.
Parallelogram with equal diagonals is a?

From the bank · past-year question

Example 5VectorsHARD

Let

P,Q,R

and

S

be the points on the plane with position vectors

-2\hat{i}-\hat{j},\, 4\hat{i},\, 3\hat{i}+3\hat{j}

and

-3\hat{i}+2\hat{j}

respectively. Then the quadrilateral PQRS must be a

[Q102 · 15th May Shift 2 · 2023]

Equal diagonals $\to$ rectangle, perpendicular diagonals $\to$ rhombus

Don't mix the two tests. A parallelogram whose diagonals are EQUAL in length is a rectangle; one whose diagonals are PERPENDICULAR is a rhombus. A figure that is a parallelogram but neither (diagonals unequal AND not perpendicular) is the 'neither rhombus nor rectangle' answer.

Right-angle test needs side-vectors FROM the vertex

For a right angle at

A

, dot

\vec{AB}

with

\vec{AC}

(both leaving

A

) — not

\vec{AB}

with

\vec{BC}

. Using the wrong pair tests perpendicularity at the wrong vertex and gives a spurious value.

Drill 1 more on triangle and parallelogram applications

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 (5)

Section formula — internal, external, and midpoint
Section formula (internal / external / midpoint)
$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n} \qquad \vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n} \qquad \vec{r} = \frac{\vec{a} + \vec{b}}{2}$
Centroid and median identities
Centroid and median vector
$\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \qquad \vec{AD} = \frac{\vec{AB} + \vec{AC}}{2} \qquad \vec{g}_{\text{tetra}} = \frac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4}$
Finding the ratio, collinearity, and cevian intersection
Ratio recovery and external division
$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n} \;\Rightarrow\; \text{ratio } m:n \qquad \vec{r}_{\text{ext}} = \frac{m\vec{b} - n\vec{a}}{m - n}$
Incentre, orthocentre, and the angle bisector
Incentre and angle bisector
$\vec{I} = \frac{a\vec{a} + b\vec{b} + c\vec{c}}{a + b + c} \qquad \text{bisector of }\angle AOB \parallel \frac{\vec{a}}{|\vec{a}|} + \frac{\vec{b}}{|\vec{b}|}$
Triangle and parallelogram applications
Right-angle test and parallelogram diagonals
$\angle A = 90^\circ \iff \vec{AB}\cdot\vec{AC} = 0 \qquad \vec{PR} = \vec{PQ} + \vec{QR}, \;\; \vec{QS} = \vec{QP} + \vec{PS}$

Watch out for (11)

Internal adds, external subtracts
→ Section formula — internal, external, and midpoint
Which point gets the weight $m$ ?
→ Section formula — internal, external, and midpoint
Median length $\neq$ half the side it bisects
→ Centroid and median identities
Centroid uses position vectors, not side vectors
→ Centroid and median identities
Internal and external points use the SAME magnitude of ratio
→ Finding the ratio, collinearity, and cevian intersection
Cevian-intersection ratio is asked along ONE cevian
→ Finding the ratio, collinearity, and cevian intersection
Incentre weights are OPPOSITE side lengths
→ Incentre, orthocentre, and the angle bisector
Bisector uses unit vectors — sum, not difference
→ Incentre, orthocentre, and the angle bisector
Orthocentre $\neq$ centroid $\neq$ circumcentre
→ Incentre, orthocentre, and the angle bisector
Equal diagonals $\to$ rectangle, perpendicular diagonals $\to$ rhombus
→ Triangle and parallelogram applications
Right-angle test needs side-vectors FROM the vertex
→ Triangle and parallelogram applications

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1VectorsMODERATE

If the vectors

\overrightarrow{AB} = 3\hat{i} + 4\hat{k}

and

\overrightarrow{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}

are the sides of the triangle ABC, then the length of the median, through A, is

[Q129 · 16th May Shift 1 · 2023]

Example 2VectorsHARD

Let

\vec{p}

and

\vec{q}

be the position vectors of P and Q respectively, with respect to O and

|\vec{p}|=p

|\vec{q}|=q

. The points R and S divide PQ internally and externally in the ratio 2:3 respectively. If OR and OS are perpendiculars, then

[Q119 · 4th May Shift 1 · 2023]

Example 3VectorsHARD

The incentre of the triangle ABC, whose vertices are

A(0,2,1), B(-2,0,0)

and

C(-2,0,2)

[Q111 · 4th May Shift 2 · 2023]

Example 4VectorsMODERATE

\triangle ABC

is right angled at

A

, where

A=(4,2,x)

B=(3,1,8)

and

C\equiv(2,-1,2)

, then the value of

x

is:

[Q140 · 14th May Shift 1 · 2024]

Example 5VectorsMODERATE

If the vectors

\overrightarrow{AB}=3\hat{i}+4\hat{k}

and

\overrightarrow{AC}=5\hat{i}-2\hat{j}+4\hat{k}

are the sides of the triangle ABC, then the length of the median through A is

[Q122 · 3rd May Shift 2 · 2023]

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