SECTION FORMULA

Let A and B be two towns. Assume that one can reach town B from A by moving 60 km towards east and then 30 km towards north. A telephone company wants to raise a relay tower at P which divides the line joining A and B in the ratio 1 : 2 internally.

Now, it wants to find the position of P where the relay tower is to be set up. Choose the point A as the origin. Let P(x, y) be the point. Draw the perpendiculars from P and B to the x-axis, meeting it in C and D respectively. Also draw a perpendicular from P to BD, intersecting at E.

Since ΔPAC and ΔBPE are similar, we have

Therefore, the position of the relay tower is at P(20, 10).

Taking the above problem as a model, we shall derive the general section formula.

Let A(x₁, y₁) and B(x₂, y₂) be two distinct points such that a point P(x, y) divides AB internally in the ratio l : m.

That is,

AP/PB = l/m

From the diagram above, we get

AF = CD = OD - DC = x - x₁

PG = DE = OE - OD = x₂ - x

Also,

PF = PD - FD = y - y₁

BG = BE - GE = y₂ - y

Now, ΔAFP and ΔPGB are similar.

Thus

Thus, the point P which divides the line segment joining the two points A(x₁, y₁) and B(x₂, y₂) internally in the ratio l : m is

This formula is known as section formula.

It is clear that the section formula can be used only when the related three points are collinear.

Results

Let AB be a line segment joining the two points A(x₁, y₁) and B(x₂, y₂) externally in the ratio l : m, then the point P is :

Solved Problems

Problem 1 :

Find the point which divides the line segment joining the points (3 , 5) and (8 , 10) internally in the ratio 2 : 3.

Solution :

Let A(3, 5) and B(8, 10) be the given points.

Let the point P(x, y) divide the line AB internally in the ratio 2 : 3.

By section formula,

Here (x₁, y₁) = (3, 5), (x₂, y₂) = (8, 10), l = 2 and m = 3.

= [2(8) + 3(3)] / (2 + 3), [2(10) + 3(5)] / (2 + 3)

= (16 + 9)/5, (20 + 15)/5

= 25/5, 35/5

= (5, 7)

So, the required point is (5, 7)

Problem 2 :

In what ratio does the point P(-2, 3) divide the line segment joining the points A(-3, 5) and B(4, -9) internally?

Solution :

Given points are A(-3, 5) and B(4, -9).

Let P (-2, 3) divide AB internally in the ratio l : m.

By the section formula,

Here (x₁, y₁) = (-3, 5) and (x₂, y₂) = (4, -9).

Equating the x-coordinates, we get

Hence P divides AB internally in the ratio 1 : 6.

Problem 3 :

Find the points of trisection of the line segment joining

(4, -1) and (-2, -3)

Solution :

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

Let P(x, y) and Q(a, b) be the points of trisection of AB so that

AP = PQ = QB

Hence P divides AB internally in the ratio 1 : 2 and Q divides AB internally in the ratio 2 : 1.

By the section formula, the required points are

Note that Q is the midpoint of PB and P is the midpoint of AQ.

Problem 4 :

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

Solution :

Let A(3, 4) and B(-6, 2) be the given points.

Let the point P(x, y) divide the line AB externally in the ratio 3 : 2.

By section formula,

Here (x₁, y₁) = (3, 4), (x₂, y₂) = (-6, 2), l = 3 and m = 2.

Problem 5 :

Find the points which divide the line segment joining (-4, 0) and (0, 6) into four equal parts.

Solution :

Let A(-4, 0) and B(0, 6) be the given points.

Let P, Q and R be the three points which divide the line AB into four equal parts.

P divides the line segment in the ratio 1 : 3.

By section formula, point P :

Q divides the line segment in the ratio 2 : 2.

By section formula, point Q :

R divides the line segment in the ratio 3 : 1.

By section formula, point R :

Problem 6 :

A (2, 5), B (-1, 2) and C (5, 8) are vertices of triangle ABC. P and Q are points on AB and AC respectively such that AP : PB = AQ : QC = 1 : 2

a) Find the cooridnates of points P and Q.

b)  Show that BC = 3PQ

Solution :

A (2, 5), B (-1, 2) and C (5, 8)

AB, BC and CA are the sides. P is the point on the side of triangle AB and divides the line segment A and B at the ratio of 1 : 2.

a)  

Finding the point P :

= (mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)

= (1(-1) + 2(2))/(1 + 2), (1(2) + 2(5))/(1 + 2)

= (-1 + 4)/3, (2 + 10)/3

= (3/3, 12/3)

= (1, 4)

Finding the point Q :

= (1(5) + 2(2))/(1 + 2), (1(8) + 2(5))/(1 + 2)

= (5 + 4)/3, (8 + 10)/3

= (9/3, 18/3)

= (3, 6)

b)

Distance between the points B and C :

B (-1, 2) and C (5, 8)

= √(x₂ - x₁)² + (y₂ - y₁)²

= √(5 + 1)² + (8 - 2)²

= √6² + 6²

= √(36 + 36)

= √72

= 6√2

Distance between the points P and Q :

P(1, 4) Q(3, 6)

= √(3 - 1)² + (6 - 4)²

= √2² + 2²

= √(4 + 4)

= √8

= 2√2

6√2 = 3(2√2)

Hence it is proved.

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