**Section Formula**

Let A and B be two points in the plane of the paper as shown in fig. and P be a point on the segment joining A and B such that AP : BP = m : n. Then, the point P divides segment AB internally in the ratio m : n.

If P is a point on AB produced such that AP : BP = m : n, then point P is said to divide AB externally in the ratio m : n.

The coordinates of the point which divides the line segment joining the points (x_{1}, y_{1}) and (x_{2}, y_{2}) internally in the ratio m : n are given by

The coordinates of P are

**Note 1:**

If P is the mid-point of AB, then it divides AB in the ratio 1 : 1, so its coordinates are

**Note 2:**

Fig. will help to remember the section formula.

**Note 3:**

The ratio m : n can also be written as or λ : 1, where λ =

So, the coordinates of point P dividing the line segment joining the points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) are

Read More: Distance between two points

**Section Formula With Examples**

**Type I: On finding the section point when the section ratio is given**

**Example 1:** 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.

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

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

**Example 2:** Find the coordinates of points which trisect the line segment joining (1, –2) and (–3, 4).

**Sol.** Let A(1, –2) and B(–3, 4) be the given points.

Let the points of trisection be P and Q. Then,

AP = PQ = QB = λ (say).

∴ PB = PQ + QB = 2λ and AQ = AP + PQ = 2λ

⇒ AP : PB = λ : 2λ = 1 : 2 and

AQ : QB = 2λ : λ = 2 : 1

So, P divides AB internally in the ratio 1 : 2 while Q divides internally in the ratio 2 : 1. Thus, the coordinates of P and Q are

Hence, the two points of trisection are (–1/3, 0) and (–5/3, 2).

**Type II: On Finding the section ratio or an end point of the segment when the section point is given**

**Example 3:** 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.

**Sol.** Let the required ratio be λ : 1. Then, the coordinates of the point of division are,

But, it is a point on x-axis on which y-coordinates of every point is zero.

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

**Example 4:** If the point C (–1, 2) divides internally the line segment joining A (2, 5) and B in ratio 3 : 4, find the coordinates of B.

**Sol.** Let the coordinates of B be (α, β). It is given that AC : BC = 3 : 4. So, the coordinates of C are

But, the coordinates of C are (–1, 2)

⇒ α = – 5 and β = – 2

Thus, the coordinates of B are (–5, –2).

**Example 5:** Determine the ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7).

**Sol. ** Suppose the line 3x + y – 9 = 0 divides the line segment joining A (1, 3) and B(2, 7) in the ratio k : 1 at point C. Then, the coordinates of C are

But, C lies on 3x + y – 9 = 0. Therefore,

⇒ 6k + 3 + 7k + 3 – 9k – 9 = 0

⇒ k =

So, the required ratio is 3 : 4 internally.

**Type III : On determination of the type of a given quadrilateral**

**Example 6:** Prove that the points (–2, –1), (1, 0), (4, 3) and (1, 2) are the vertices of a parallelogram. Is it a rectangle ?

**Sol.** Let the given point be A, B, C and D respectively. Then,

Coordinates of the mid-point of AC are

Coordinates of the mid-point of BD are

Thus, AC and BD have the same mid-point. Hence, ABCD is a parallelogram.

Now, we shall see whether ABCD is a rectangle or not.

We have,

Clearly, AC ≠ BD. So, ABCD is not a rectangle.

**Example 7:** Prove that (4, – 1), (6, 0), (7, 2) and (5, 1) are the vertices of a rhombus. Is it a square?

**Sol. ** Let the given points be A, B, C and D respectively. Then,

Coordinates of the mid-point of AC are

Coordinates of the mid-point of BD are

Thus, AC and BD have the same mid-point.

Hence, ABCD is a parallelogram.

∴ AB = BC

So, ABCD is a parallelogram whose adjacent sides are equal.

Hence, ABCD is a rhombus.

We have,

Clearly, AC ≠ BD.

So, ABCD is not a square.

**Type IV: On finding the unknown vertex from given points**

**Example 8:** 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.

**Sol.** 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.

∴ Coordiantes of the mid-point of AC = Coordinates of the mid-point of BD

⇒ x = – 2 and y = 1

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

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

**Sol. ** We know that the diagonals of a parallelogram bisect each other. So, coordinates of the mid-point of diagonal AC are same as the coordinates of the mid-point of diagonal BD.

⇒ 15 = 8 + p

⇒ p = 7

**Example 10:** If A(–2, –1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b.

**Sol. ** We know that the diagonals of a parallelogram bisect each other. Therefore, the coordinates of the mid-point of AC are same as the coordinates of the mid-point of BD i.e.,

⇒ a + 1 = 2 and b – 1 = 2

⇒ a = 1 and b = 3

**Example 11:** If the coordinates of the mid-points of the sides of a triangle are (1, 2) (0, –1) and (2, 1). Find the coordinates of its vertices.

**Sol.** Let A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) be the vertices of ∆ABC. Let D (1, 2), E (0, –1), and F(2, –1) be the mid-points of sides BC, CA and AB respectively. Since D is the mid-point of BC.

⇒ x_{2} + x_{3} = 2 and y_{2} + y_{3} = 4 …. (1)

Similarly, E and F are the mid-point of CA and AB respectively.

⇒ x_{1} + x_{3} = 0 and y_{1} + y_{3} = – 2 …. (2)

⇒ x_{1} + x_{2} = 4 and y1 + y2 = –2 …. (3)

From (1), (2) and (3), 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

⇒ 2(x_{1} + x_{2} + x_{3}) = 6 and 2(y_{1} + y_{2} + y_{3}) = 0 …. (4)

⇒ x_{1} + x_{2} + x_{3} = 3

and y_{1} + y_{2} + y_{3} = 0

From (1) and (4), we get

x_{1} + 2 = 3 and y_{1} + 4 = 0

⇒ x_{1} = 1 and y_{1} = – 4

So, the coordinates of A are (1, – 4)

From (2) and (4), we get

x_{2} + 0 = 3 and y_{2} – 2 = 0

⇒ x_{2} = 3 and y_{2} = 2

So, coordinates of B are (3, 2)

From (3) and (4), we get

x_{3} + 4 = 3 and y_{3} – 2 = 0

⇒ 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).

**Example 12:** Find the lengths of the medians of a ∆ABC whose vertices are A(7, –3), B(5,3) and C(3,–1).

**Sol. ** Let D, E, F be the mid-points of the sides BC, CA and AB respectively. Then, the coordinates of D, E and F are

**Example 13:** If A (5, –1), B(–3, –2) and C(–1, 8) are the vertices of triangle ABC, find the length of median through A and the coordinates of the centroid.

**Sol.** Let AD be the median through the vertex A of ∆ABC. Then, D is the mid-point of BC. So, the coordinates of

Let G be the centroid of ∆ABC. Then, G lies on median AD and divides it in the ratio 2 : 1. So, coordinates of G are

**Application Of Section Formula**

The coordinates of the centroid of the triangle whose vertices are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are

**Example 14:** Find the coordinates of the centroid of a triangle whose vertices are (–1, 0), (5, –2) and (8, 2).

**Sol.** We know that the coordinates of the centroid of a triangle whose angular points are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are

So, the coordiantes of the centroid of a triangle whose vertices are (–1, 0), (5, –2) and (8, 2) are

**Example 15:** If the coordinates of the mid points of the sides of a triangle are (1, 1), (2, – 3) and (3, 4) Find its centroid.

**Sol.** Let P (1, 1), Q(2, –3), R(3, 4) be the mid-points of sides AB, BC and CA respectively of triangle ABC. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) be the vertices of triangle ABC. Then, P is the mid-point of BC

⇒ x_{1} + x_{2} = 2 and y_{1} + y_{2} = 2 …(1)

Q is the mid-point of BC

⇒ x_{2} + x_{3} = 4 and y_{2} + y_{3} = – 6 …(2)

R is the mid-point of AC

⇒ x_{1} + x_{3} = 6 and y1_{1} + y_{3} = 8 …(3)

From (1), (2) and (3), we get

x_{1} + x_{2} + x_{2} + x_{3} + x_{1} + x_{3} = 2 + 4 + 6

and, y_{1} + y_{2} + y_{2} + y_{3} + y_{1} + y_{3} = 2 – 6 + 8

x_{1} + x_{2} + x_{3} = 6 and y_{1} + y_{2} + y_{3} = 2 …(4)

The coordinates of the centroid of ∆ABC are

**Example 16:** Two vertices of a triangle are (3, –5) and (–7, 4). If its centroid is (2, –1). Find the third vertex.

**Sol. ** Let the coordinates of the third vertex be (x, y). Then,

⇒ x – 4 = 6 and y – 1 = – 3

⇒ x = 10 and y = – 2

Thus, the coordinates of the third vertex are (10, –2).

**Example 17:** Prove that the diagonals of a rectangle bisect each other and are equal.

**Sol. ** Let OACB be a rectangle such that OA is along x-axis and OB is along y-axis. Let OA = a and OB = b.

Then, the coordinates of A and B are (a, 0) and (0, b) respectively.

Since, OACB is a rectangle. Therefore,

AC = Ob ⇒ AC = b

Thus, we have

OA = a and AC = b

So, the coordiantes of C are (a, b).

The coordinates of the mid-point of OC are

Also, the coordinates of the mid-points of AB are

Clearly, coordinates of the mid-point of OC and AB are same.

Hence, OC and AB bisect each other.

∴ OC = AB

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