## Section formula

Section formula is used to determine the coordinate of a point that divides a line into two parts such that ratio of their length is m:n.

Let P and Q be the given two points (x_{1},y_{1}) and (x_{2},y_{2}) respectively,

and M be the point dividing the line-segment PQ internally in the ratio m:n,

then form the sectional formula for determining the coordinate for a point M is given as:

**\(\large M (x,y) = \left ( \frac{mx_{2}+nx_{1}}{m+n} , \frac{my_{2}+ny_{1}}{m+n} \right )\)**

### Proof for Sectional Formula:

Let \(P(x_1, y_1)\)

\(PA, ~MN ~and~ QR\)

\(PS\)

\(∠MPS\)

\(∠MSP\)

By \(AA\)

\(∆PMS\)

\(⇒~\frac{PM}{MQ}\)

\(PS\)

\(MB\)

\(MS\)

\(QB\)

Equation (2) gives

\(\large \frac{m}{n}\)

\(\large ⇒~\frac{m}{n}\)

\(\large⇒~x\)

Similarly,

\(\large\frac{m}{n}\)

\(\large⇒~y\)

So, the coordinates of the point \(M(x,y)\)

\(\LARGE\left(\frac{mx_2~+~nx_1}{m~+~n},\frac{my_2~+~ny_1}{m~+~n}\right)\)

This is known as **section formula**.

### Sectional formula (Externally):

Sectional Formula can also be used to find the coordinate of a point that lie outside the line, where the ratio of length of a point from both the lines segments are in the ratio m:n.

The Sectional Formula is given as:** \(\large \left ( \frac{mx_{2} – nx_{1}}{m – n} , \frac{my_{2} – ny_{1}}{m – n} \right )\)**

**Let us understand it’s proof with an Example:**

Consider a point \(A(x_1, y_1). C(x,y)\)

\(⇒~OC:CA\)

To find coordinates of the point \(C\)

\(∠COD\)

\(∠CDO\)

By \(AA\)

\(∆OCD\)

So, the ratio of corresponding sides will be equal, i.e.

\(\frac{OC}{CA}\)

Since coordinates of \(C\)

\(OD\)

\(BE\)

\(CE\)

\(AE\)

Also, \(OC:CA\)

\(⇒~OC\)

Equation(1) becomes

\(\large \frac{2a}{a}\)

\(\large \frac{x}{x_1~-~x}\)

\(\large ⇒~x\)

\(\large ⇒~3x\)

\(\large ⇒~x\)

Similarly,

\(\large \frac{y}{y_1~-~y}\)

\(\large ⇒~3y\)

\(\large y\)

Therefore, the coordinates of the point \(C\)

In general, the coordinates of a point which divides the line joining points with coordinates (\(x_1, y_1\)

If the point \(M\)

\(\large \left(\frac{kx_2~+~x_1}{k~+~1},\frac{ky_2~+~y_1}{k~+~1}\right)\)

What if the point \(M\)**line segment** joining points \(P(x_1, y_1)\)

If \(M\)

Coordinates of point \(M\)

\(\large \left(\frac{1~×~x_2~+~1~×~x_1}{1~+~1}, \frac{1~×~y_2~+~1~×~y_1}{1~+~1}\right)\)

Therefore, **coordinates of a point** which is the midpoint of line segment joining points \((x_1,y_1)\)

\(\large \left(\frac{x_1~+~x_2}{2}, \frac{y_1~+~y_2}{2}\right)\)

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

** Solution: **Let \(P(x,y)\)

\(\large \left(\frac{mx_2~+~nx_1}{m~+~n}, \frac{my_2~+~ny_1}{m~+~n}\right)\)

\(m\)

Coordinates of \(P\)

\(x\)

\(⇒~x\)

\(⇒~x\)

*Example:* The 4 vertices of a parallelogram are \(A\)

* Solution:* We know that diagonals of a parallelogram bisect each other.

Let \(O\)

Coordinates of mid-points of both \(\overleftrightarrow{AC}\)

Therefore,

Using midpoint section formula,

\(\large \left(\frac{x_1~+~x_2}{2}, \frac{y_1~+~y_2}{2}\right)\)

\(\large \frac{-2~+~p}{2}\)

\(\large \Rightarrow ~-2~+~p\)

Similarly,

\(\large\frac{3~+~q}{2}\)

\(3~+~q\)

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