## Coordinate Geometry | Section Formula

### Section formula :

The coordinates of the point *P*(*x*, *y*) which divides the line segment joining and internally in the ratio m : n are given by

**Proof: **

Let *X’OX* and *YOY’* be the coordinate axes.

Let and be the end points of the given line segment *AB*.

Let *P*(*x*, *y*) be the point which divides *AB* in the ratio *m* : *n.*

Then,

Draw *AL *^ *OX*; *BM *^ *OX*; *PN *^ *OX*; *AR *^ *PN*; and *PS *^ *BM*

Now, *AR* = *LN* = *ON* – *OL* =

*PS* = *NM* = *OM* – *ON* =

*PR* = *PN* – *RN* = *PN* – *AL* =

*BS* = *MB* – *SM* = *BM* – *PN* = .

D*ARP* and D*PSB* are similar and, therefore, their sides are proportional.

\

Þ

Þ

Þ and

Þ and

Þ

Hence, the coordinates of *P* are .

### Midpoint formula :

The coordinates of the midpoint *M* of a line segment *AB* with end points and are .

**Proof:*** *

* *Let M be the midpoint of the line segment joining the points and .

Then, M divides AB in the ratio 1 : 1

So, by the section formula, the coordinates of M are

, i.e., .

Hence, the coordinates of the midpoint of AB are .

### Question:

#### Find the coordinates of the point which divides the line segment joining the points (6, 3) and (–4, 5) in the ratio 3 : 2 internally.

### Solution:

Let *P*(x, y) be the required point. Then,

and

Þ *x* = 0 and *y* =

So, the coordinates of *P* are (0, 21/5).

### Question:

#### In what ratio does the point* C*(3/5, 11/5) divide the line segment joining the points* A*(3, 5) and* B*(–3, –2)?.

### Solution:

Let the point *C* divide *AB* in the ratio *k* : 1. Then the coordinates of *C* are

But, the coordinates of *C* are given as (3/5, 11/5).

\ and

Þ – 15*k* + 15 = 3*k* + 3 and – 10*k* + 25 = 11*k* + 11

Þ 18*k* = 12 and 21*k* = 14

Þ . Hence, the point C divides AB in the ratio 2 : 3.

### Question:

#### Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points* A*(2, –2) and *B*(–7, 4).

### Solution:

Let *P* and *Q* be the points of trisection of *AB* i.e., *AP* = *PQ* = *QB*.

Therefore, *P* divides *AB* internally in the ratio 1 : 2. Therefore, the coordinates of *P*, by applying the section formula, are

, i.e., (–1, 0)

Now, *Q* also divides *AB* internally in the ratio 2 : 1. So, the coordinates of *Q* are

, i.e., (–4, 2)

Therefore, the coordinates of the points of trisection of the line segment joining *A* and *B* are (–1, 0) and (–4, 2).

### Question:

#### Find the ratio in which the y-axis divides the line segment joining the points (5, –6) and (–1, –4). Also find the point of intersection.

### Solution:

Let the ratio be *k* : 1. Then by the section formula, the coordinates of the point which divides *AB* in the ratio *k* : 1 are .

This point lies on the *y*-axis, \ its coordinates are (0, *y*)

\

So, *k* = 5

Thus, the ratio is 5 : 1.

Putting the value of *k* = 5, we get the point of intersection as .

### Question:

#### In what ratio does the x-axis divide the line segment joining the points (2, –3) and (5, 6)? Also, find the coordinates of the point of intersection.

### Solution:

Let the required ratio be *k* : 1. Then, the coordinates of the point *R* which divides *PQ *in the ratio *k *: 1 are

This point lies on x-axis therefore its coordinates are (*x*, 0)

\ . Thus, the required ratio is or 1 : 2.

Putting *k* = 1/2 in the coordinates of *R*, we find that its coordinates are (3, 0).

### Question:

#### If the points* A*(6, 1),* B*(8, 2),* C*(9, 4) and* D*(*p*, 3) are the vertices of a parallelogram, taken in order, find the value of* p.*

### Solution:

In parallelogram *ABCD*

coordinates of the mid-point of *AC* = coordinates of the mid-point of *BD*

Þ

Þ

Þ

Þ 30 = 16 + 2*p*

Þ 14 = 2*p*

Þ *p* = 7

### Question:

#### The three vertices of a parallelogram taken in order are (–1, 0), (3, 1) and (2, 2) respectively. Find the coordinates of the fourth vertex.

### Solution:

Let *A*(–1, 0), *B*(3, 1), *C*(2, 2) and *D*(x, y) be the vertices of a parallelogram *ABCD* taken in order. Since, the diagonals of a parallelogram bisect each other.

\ Coordinates of the mid-point of *AC* = Coordinates of the mid-point of *BD
*

Þ

Þ

Þ and

Þ *x* = – 2 and *y* = 1

Hence, the fourth vertex of the parallelogram is (–2, 1).

### Question:

#### Find the ratio in which the point (–3,* p*) divides the line segment joining the points (–5, –4) and (–2, 3). Hence, find the value of* p.*

### Solution:

Suppose the point *P*(–3, *p*) divides the line segment joining points *A*(–5, –4) and *B*(–2, 3) in the ratio *k* : 1.

Then, the coordinates of *P* are

But, the coordinates of *P* are given as (–3, *p*).

\ and

Þ –2*k* – 5 = –3*k* – 3 and

Þ *k* = 2 and *p* =

Þ *k* = 2 and *p* = .

Hence, the ratio is 2 : 1 and *p* = .

### Question:

#### The coordinates of one end point of a diameter *AB *of a circle are* A*(4, –1) and the coordinates of the centre of the circle are *C*(1, –3). Find the coordinates of *B.*

### Solution:

In the figure *A*(4, –1) and *B*(*a*, *b*) are the end points of the given diameter *AB*, and *C*(1, –3) is the centre of the circle. Then, *C* is the midpoint of *AB*.

By the midpoint formula, the coordinates of *C* are .

But, the coordinates of *C* are given as (1, –3).

\

Þ 4 + *a* = 2 and –1 + *b* = –6

Þ *a* = –2 and *b* = –5

Hence, the required point is *B*(–2, –5).

### Question:

#### If *A*(5, –1),* B*(–3, –2) and* C*(–1, 8) are the vertices of triangle *ABC*, find the length of median through *A*.

### Solution:

Let *AD* be the median through the vertex *A* of D*ABC*. Then, *D* is the midpoint of *BC*.

By midpoint formula, the coordinates of *D* are i.e., (–2, 3).

\

units

### Question:

#### If the coordinates of the midpoints of the sides of a triangle are (1, 2)(0, –1) and (2, –1). Find the coordinates of its vertices.

### Solution:

Let *A*(x_{1}, *y*_{1}), *B*(*x*_{2}, *y*_{2}) and *C*(*x*_{3}, *y*_{3}) be the vertices of D*ABC*. Let *D* (1, 2), *E*(0, –1), and *F*(2, –1) be the midpoints of sides *BC*, *CA* and *AB* respectively. Since *D* is the midpoint of *BC*.

\ and

Þ *x*_{2} + *x*_{3} = 2 and *y*_{2} + *y*_{3} = 4 …(i)

Similarly, *E* and *F* are midpoints of *CA* and *AB* respectively.

\ and

Þ *x*_{1} + *x*_{3} = 0 and *y*_{1} + *y*_{3} = – 2 …(ii)

and, and

Þ *x*_{1} + *x*_{2} = 4 and *y*_{1} + *y*_{2} = – 2 …(iii)

From (i), (ii) and (iii), we get

(*x*_{2} + *x*_{3}) + (*x*_{1} + *x*_{3}) + (*x*_{1} + *x*_{2}) = 2 + 0 + 4 and

(*y*_{2} + *y*_{3}) + (*y*_{1} + *y*_{3}) + (*y*_{1} + *y*_{2}) = 4 – 2 – 2

Þ 2(*x*_{1} + *x*_{2} + *x*_{3}) = 6 and 2(*y*_{1} + *y*_{2} + *y*_{3}) = 0

Þ *x*_{1} + *x*_{2} + *x*_{3} = 3 and *y*_{1} + *y*_{2} + *y*_{3} = 0 …(iv)

Þ From (i) and (iv), we get

*x*_{1} + 2 = 3 and *y*_{1} + 4 = 0

Þ *x*_{1} = 1 and *y*_{1} = – 4. So, the coordinates of A are (1, – 4)

From (ii) and (iv), we get

*x*_{2} + 0 = 3 and *y*_{2} – 2 = 0

Þ *x*_{2} = 3 and *y*_{2} = 2. So, coordinates of *B* are (3, 2).

From (iii) and (iv), we get

*x*_{3} + 4 = 3 and *y*_{3} – 2 = 0

Þ *x*_{3} = – 1 and *y*_{3} = 2. So, coordinates of *C* are (–1, 2).

Hence, the vertices of the triangle *ABC* are *A*(1, – 4), *B*(3, 2) and *C*(–1, 2).

### Centroid of a Triangle

The coordinates of the centroid of a triangle with vertices and is given by .

**Proof:
**

Let and

be the vertices of a D*ABC*.

Let *D* be the midpoint of *BC*

Then, the coordinates of *D* by midpoint formula are .

Let *G*(*x*, *y*) be the centroid of D*ABC*.

Then, *G* divides *AD* in the ratio 2 : 1

\

Hence, the coordinates of *G* are .

### Question:

#### Find the centroid of *DABC *whose vertices are* A*(–3, 0),* B*(5, –2) and* C*(–8, 5).

### Solution:

Here, and

Let *G*(*x*, *y*) be the centroid of D*ABC*. Then,

Hence, the centroid of D*ABC* is *G*(–2, 1)

### Question:

#### Two vertices of a D*ABC* are given by* A*(6, 4) and* B*(–2, 2), and its centroid is* G*(3, 4). Find the coordinates of the third vertex* C* of D*ABC.*

### Solution:

Two vertices of D*ABC* are *A*(6, 4) and *B*(–2, 2). Let the third vertex by *C*(*a*, *b*).

Then, the coordinates of its centroid are

, i.e.,

But, it is given that the centroid is *G*(3, 4).

\ and

Þ *a* = 5 and *b* = 6

Hence, the third vertex of D*ABC* is *C*(5, 6)