Section Formula: Definition, Formula, Derivation and Examples

Section Formula: If you want to reach our office, we will give our office address such as building number, street name, and area code. These are the references to get our office. Similarly, how will we locate a point in a two-dimensional space? For that, we take the help of coordinate geometry.

The values of the \(x\)-axis and \(y\)-axis are the references for locating a point in the coordinate plane or a plane surface. The section formula tells us the coordinates of the point, which divides the given line segment into two parts such that their lengths are in a particular ratio.

What is Section Formula?

Section Formula is a coordinate point that divides a line segment internally or externally. Section Formula is used to divide a line segment into any ratio or to find the midpoint between the two ends of the line segment.

CLEAR YOUR CONCEPTUAL DOUBTS ON SECTION FORMULA

The study of geometry using coordinate points or coordinate systems is known as coordinate geometry. It describes the relationship between geometry and algebra through graphs involving curves and lines.

Coordinate Axes

The two reference axes, \(x\)-axis, and \(y\)-axis are called coordinate axes.
The combination of coordinates and coordinate axes is called a coordinate system and this system is introduced by the great French mathematician Rene Descartes.
Therefore, this coordinate system is named after him as a Cartesian coordinate system.

Coordinates

In the coordinate plane, the \(x\) and \(y\) axis divide the plane into four parts, or we call it as quadrants. The meeting point of the two coordinate axes is called the origin, \(O\). The right side of the origin in the horizontal line possesses the positive values or it is known as the positive \(x\) axis and the left side of the origin possesses the negative values, or it is known as the negative \(x\) axis. Similarly in the vertical line, positive values of \(y\) axis are above the origin and negative values of \(y\) axis are below the origin. These values of \(x\) and \(y\) axis help us to locate any point on the surface of the plane.
For example: To find the location of point \(P\) in the below-given plane, we need references.

We will take reference of \(x\)-axis and \(y\)-axis. In the given plane, the positive \(x\)-axis is covering \(2\) units and the positive \(y\)-axis is covering \(3\) units.

This kind of representation of the location of the point is called coordinates. The value of the \(x\)-axis is known as \(x\)-coordinate or abscissa and the value of the \(y\)-axis is known as \(y\)-coordinate or ordinate.

Sign of the Coordinates

The sign of the coordinates is completely dependent on the quadrant in which they exist. The reference axis that is the \(x\)-axis and \(y\)-axis is called coordinate axes.
The four types of quadrants are:
1. First quadrant
2. Second quadrant
3. Third quadrant
4. Fourth quadrant
In the first quadrant, both \(x\) and \(y\) values are positive.
In the second quadrant, \(x\) is negative and \(y\) is positive.
In the third quadrant, both \(x\) and \(y\) values are negative.
In the fourth quadrant, \(x\) is positive and \(y\) is negative.

Graphical Method to Find the Coordinates of a Point

Let us consider two houses, house \(A\) and house \(B\). The coordinates give the locations of the houses. Let us say the coordinates of \(A\) are \(\left( {2,\,3} \right)\) and the coordinates of \(B\) are \(\left( {5,\,6} \right)\). The city development authority needs to place a light post at a point \(P\) anywhere in between \(A\) and \(B\), such that it divides the line joining \(A\) and \(B\) into two sections and those sections are such a way that the distance between \(A\) and \(P\) is twice the distance between \(P\) and \(B\).

It is represented as,
\(\frac{{AP}}{{PB}} = \frac{2}{1}\)
Or we can say that the ratio of their distance is \(AP:PB = 2:1\)
This indicates that if the distance between \(A\) and \(B\) is divided into three equal parts, then \(2\) parts will be on the left side of \(P\) that is towards \(A\), whereas one part is on the right side of the \(P\) that is towards \(B\).

We need to find the coordinates of \(P\) that is \(x\) and \(y\).
Let us start with \(x\) coordinate. \(x\) coordinate of \(A\) is \(2\) and \(x\)-coordinate of \(B\) is \(5\). Looking at the ratio \(2:1\), we realize that we need to find the value of \(x\) in such a way that if the distance between \(2\) and \(5\) is divided into three equal parts, then two of the parts will be towards point \(2\) that is \(x\)coordinate of \(A\) and one part will be towards \(5\) that is \(x\) coordinate of \(B\). This can only be done by point \(4\) on the \(x\)-axis. So, the value of \(x\)-coordinate of \(P\) is \(4\).

Now, we can see that the distance between \(2\) and \(5\) is divided in the ratio \(2:1\).
Now, let us consider the ordinate of \(P\). The ordinate of \(A\) is \(3\) and the ordinate of \(B\) is \(6\) and we can see that point \(5\) on the positive \(y\)-axis satisfies the ratio \(2:1\).

So, therefore, the value of the \(y\) coordinate of \(P\) is \(5\).
Now, we got the values of coordinates of point \(P\) very quickly because the ratio values were very small, that is \(2:1\), so we have divided the total distance into \(2 + 1 = 3\) equal parts to locate the coordinates of the point \(P\). But it becomes tough to identify the points if we have big ratios. So, locating the coordinates of the point in the above graphical method for big ratios is difficult. So, to find the coordinates of the point having a big ratio, we use the section formula.

Section Formula Definition

The formula used to determine the coordinates of a point that divides the line segment joining two points into two parts with the given ratio is known as the section formula.

Derivation of Section Formula

Consider any two points \(A\left( {{x_1},\,{y_1}} \right)\) and \(B\left( {{x_2},\,{y_2}} \right)\) and assume that \(P(x,\,y)\) divides \(AB\) internally in the ratio \(m:n\) that is, \(\frac{{PA}}{{PB}} = \frac{m}{n}\).

Draw \(AR,\,PS\) and \(BT\) perpendicular to the \(x\)-axis. Draw \(AQ\) and \(PC\) parallel to the \(x\)-axis. Then, by the \(AA\) similarity criterion,
\(\Delta PAQ \sim \Delta BPC\)
Therefore, \(\frac{{PA}}{{BP}} = \frac{{AQ}}{{PC}} = \frac{{PQ}}{{BC}}\,…….(1)\)
Now,
\(AQ = RS = OS – OR = x – {x_1}\)
\(PC = ST = OT – OS = {x_2} – x\)
\(PQ = PS – QS = PS – AR = y – {y_1}\)
\(BC = BT – CT = BT – PS = {y_2} – y\)
Substituting these values in \(\left( 1 \right)\) we get
\(\frac{m}{n} = \frac{{x – {x_1}}}{{{x_2} – x}} = \frac{{y – {y_1}}}{{{y_2} – y}}\)
Taking \(\frac{m}{n} = \frac{{x – {x_1}}}{{{x_2} – {x^\prime }}}\), we get \(x = \frac{{m{x_2} + n{x_1}}}{{m + n}}\)
Similarly, \(\frac{m}{n} = \frac{{y – {y_1}}}{{{y_2} – {y^\prime }}}\), we get \(y = \frac{{m{y_2} + n{y_1}}}{{m + n}}\)
So, the coordinates of the point \(P(x,\,y)\) which divides the line segment joining the points \(A\left( {{x_1},\,{y_1}} \right)\) and \(B\left( {{x_2},\,{y_2}} \right)\) internally in the ratio \(m:n\) is
\(\left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\,\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)
This is the section formula.

Also, the coordinates of the point \(P(x,\,y)\) which divides the line segment joining the points \(A\left( {{x_1},\,{y_1}} \right)\) and \(B\left( {{x_2},\,{y_2}} \right)\) externally in the ratio \(m:n\) is
\((x,\,y) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\,\frac{{m{y_2} – n{y_1}}}{{m – n}}} \right)\)

Solved Examples – Section Formula

Q.1. Find the coordinates of the point which divides the line segment joining the points \((5,\, – 2)\) and \(\left( {4,\,3} \right)\) in the ratio \(3:1\) internally.
Ans: Let \(P(x,\,y)\) be the required point. Using section formula, we get
\((x,\,y) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\,\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)
So, \(x = \frac{{m{x_2} + n{x_1}}}{{m + n}}\) and \(y = \frac{{m{y_2} + n{y_1}}}{{m + n}}\)
\(x = \frac{{(3 \times 4) + (1 \times 5)}}{{3 + 1}}\) and \(y = \frac{{(3 \times 3) + (1 \times – 2)}}{{3 + 1}}\)

\(x = \frac{{12 + 5}}{4}\) and \(y = \frac{{9 – 2}}{4}\)

\(x = \frac{{17}}{4}\) and \(y = \frac{7}{4}\)

So, \(\left( {\frac{{17}}{4},\,\frac{7}{4}} \right)\) are the coordinates of point \(P\).

Q.2. The coordinates of point \(A\) and \(B\) are \(( – 3,\,3)\) and \((12,\,-7)\) respectively. \(P\) is a point on the line segment \(AB\) such that \(AP:PB = 2:3\). Find the coordinates of \(P\).
Ans: Let the coordinates of \(P(x,\,y)\) divides \(AB\) in the ratio \(m:n.\)
Given: \(m:n = 2:3,\,{x_1} = – 3,\,{y_1} = 3,\,{x_2} = 12,\,{y_2} = – 7,\,m = 2\), and \(n = 3\)
We know that \(P(x,\,y) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\,\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)
So, \(x = \frac{{m{x_2} + n{x_1}}}{{m + n}}\) and \(y = \frac{{m{y_2} + n{y_1}}}{{m + n}}\)
\(x = \frac{{(2 \times 12) + (3 \times – 3)}}{{2 + 3}}\) and \(y = \frac{{(2 \times – 7) + (3 \times 3)}}{{2 + 3}}\)
\(x = \frac{{24 – 9}}{5}\) and \(y = \frac{{ – 14 + 9}}{5}\)
\(x = \frac{{15}}{5}\) and \(y = \frac{{ – 5}}{5}\)
\(x = 3\) and \(y = – 1\)
Therefore, coordinates of point \(P\) are \((3,\, – 1)\).

Q.3. The line segment joining the points \(A(3,\,2)\) and \(B(5,\,1)\) is divided at the point \(P\) in the ratio \(1:2\) and it lies on the line \(3x – 18y + k = 0\). Find the value of \(k\).
Ans: Let the coordinates of \(P(x,\,y)\) divides \(AB\) in the ratio \(m:n\).
\(A(3,\,2)\) and \(B(5,\,1)\) are the given points.
Given: \(m:n = 1:2\)
We know that \(P(x,\,y) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\,\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)
So, \(x = \frac{{m{x_2} + n{x_1}}}{{m + n}}\) and \(y = \frac{{m{y_2} + n{y_1}}}{{m + n}}\)
\(x = \frac{{(1 \times 5) + (2 \times 3)}}{{1 + 2}}\) and \(y = \frac{{(1 \times 1) + (2 \times 2)}}{{1 + 2}}\)
\(x = \frac{{5 + 6}}{3}\) and \(y = \frac{{1 + 4}}{3}\)
\(x = \frac{{11}}{3}\) and \(y = \frac{5}{3}\)
Given a point, \(P\) lies online \(3x – 18y + k = 0\)

Substituting \(x\) and \(y\), we get
\(\left( {3 \times \frac{{11}}{3}} \right) – \left( {18 \times \frac{5}{3}} \right) + k = 0\)
\(11 – 30 + k = 0\)
\( – 19 + k = 0\)
\(k = 19\)
Therefore, the value of \(k = 19\)

Q.4. A point \(P\) divides the line segment joining the points \(A(3,\, – 5)\) and \(B( – 4,\,8)\) such that \(\frac{{AP}}{{PB}} = \frac{k}{1}\). If \(P\) lies on the line \(x + y = 0\), then find the value of \(k\).
Ans: Let the coordinates of \(P(x,\,y)\) divides \(AB\) in the ratio \(m:n\).
\(A(3,\, – 5)\) and \(B( – 4,\,8)\) are the given points.
Given: \(AP:PB = k:1\)
We know that \(P(x,\,y) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\,\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)
So, \(x = \frac{{m{x_2} + n{x_1}}}{{m + n}}\) and \(y = \frac{{m{y_2} + n{y_1}}}{{m + n}}\)
\(x = \frac{{(k \times – 4) + (1 \times 3)}}{{k + 1}}\) and \(y = \frac{{(k \times 8) + (1 \times – 5)}}{{k + 1}}\)
\(x = \frac{{ – 4k + 3}}{{k + 1}}\) and \(y = \frac{{8k – 5}}{{k + 1}}\)
Given a point, \(P\) lies on the line \(x + y = 0\)

Substituting \(x\) and \(y,\) we get
\(\frac{{ – 4\,k + 3}}{{k + 1}} + \frac{{8\,k – 5}}{{k + 1}} = 0\)
\( \Rightarrow \frac{{ – 4\,k + 3 + 8\,k – 5}}{{k + 1}} = 0\)
\( \Rightarrow 4\,k – 2 = 0\)
\( \Rightarrow 4\,k = 2\)
\( \Rightarrow k = \frac{2}{4}\)
\( \Rightarrow k = \frac{1}{2}\)
Therefore,\(k = \frac{1}{2}\)

Q.5. In which quadrant the point \(P\) that divides the line segment joining the points \(A(2,\, – 5)\) and \(B(5,\,2)\) internally in the ratio \(2:3\) lies?
Ans: Let the coordinates of \(P(x,\,y)\) divides \(AB\) in the ratio \(m:n\).
\(A(2,\, – 5)\) and \(B(5,\,2)\) are the given points.
Given: \(m:n = 2:3\)
We know that \(P(x,\,y) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\,\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)
So, \(x = \frac{{m{x_2} + m{x_1}}}{{m + n}}\) and \(y = \frac{{m{y_2} + m{y_1}}}{{m + n}}\)
\( \Rightarrow x = \frac{{\left( {2 \times 5} \right) + \left( {3 \times 2} \right)}}{{2 + 3}}\) and \(y = \frac{{\left( {2 \times 2} \right) + \left( {3 \times – 5} \right)}}{{2 + 3}}\)
\( \Rightarrow x = \frac{{10 + 6}}{5}\) and \(y = \frac{{4 – 15}}{5}\)
\( \Rightarrow x = \frac{{16}}{5}\) and \(y = \frac{{ – 11}}{5}\)
Therefore, coordinates of point \(P\) are \(\left( {\frac{{16}}{5},\,\frac{{ – 11}}{5}} \right)\)
So, point \(P\) lies in the \({4^{{\rm{th}}}}\) quadrant.

Summary

In the above article, we have learnt Section Formula, how to derive the sectional formula. We also learned about Coordinate Geometry, signs of coordinates, graphical method to find the coordinates of a point. We also solved some examples for better understanding of Section Formula.

Frequently Asked Questions (FAQs) – Section Formula

Q.1. What is the section formula for internal division?
Ans: Section formula for internal division is given by \((x,\,y) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\,\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)

Q.2. What is the section formula for external division?
Ans: Section formula for external division is given by \((x,\,y) = \left( {\frac{{m{x_2} – n{x_1}}}{{m – n}},\,\frac{{m{y_2} – n{y_1}}}{{m – n}}} \right)\)

Q.3. What is the section formula?
Ans:  In \(2\)-dimensional coordinate geometry, the section formula is given by \((x,\,y) = \left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\,\frac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\)

Q.4. What is the midpoint formula to find the coordinates of the mid-point?
Ans: The formula to find the coordinates of mid-point is given by \((x,\,y) = \left( {\frac{{{x_1} + {x_2}}}{2},\,\frac{{{y_1} + {y_2}}}{2}} \right)\)

Q.5. What is the distance formula in coordinate geometry?
Ans: The distance formula in coordinate geometry is to find the distance between two points \(P\left( {{x_1},\,{y_1}} \right)\) and \(Q\left( {{x_2},\,{y_2}} \right)\) is given by \(d = \sqrt {{{\left( {{x_2} – {x_1}} \right)}^2} + {{\left( {{y_2} – {y_1}} \right)}^2}} \)

SOLVE QUESTIONS ON SECTION FORMULA

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