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 \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) be two points in the \(xy\) – plane. Let \(M(x, y)\) be the point which divides line segment \(PQ\) internally in the ratio \(m:n\).

\(PA, ~MN ~and~ QR\) are drawn perpendicular to \(x\) – axis.

\(PS\) and \(MB\) are drawn parallel to \(x\) – axis.

\(âˆ MPS\) = \(âˆ QMB\) [Corresponding Angles]

\(âˆ MSP\) = \(âˆ QBM\) = \(90Â°\)

By \(AA\) similarity criterion,

\(âˆ†PMS\) ~ \(âˆ†MQB\)

\(â‡’~\frac{PM}{MQ}\) = \(\frac{PS}{MB}\) = \(\frac{MS}{QB}\) = \(\frac{m}{n}\) —(2)

\(PS\) = \(AN\) = \(ON~-~OA\) = \(x~-~x_1\)

\(MB\) = \(NR\) = \(OR~-~ON\) = \(x_2~-~x\)

\(MS\) = \(MN~-~SN\) = \(y~-~y_1\)

\(QB\) = \(RQ~-~RB\) = \(y_2~-~y\)

Equation (2) gives

\(\large \frac{m}{n}\) = \(\large \frac{x~-~x_1}{x_2~-~x}\) = \(\large \frac{y~-~y_1}{y_2~-~y}\)

\(\large â‡’~\frac{m}{n}\) = \(\large \frac{x~-~x_1}{x_2~-~x}\)

\(\largeâ‡’~x\) = \(\large\frac{mx_2~+~nx_1}{m~+~n}\)

Similarly,

\(\large\frac{m}{n}\) = \(\large\frac{y~-~y_1}{y_2~-~y}\)

\(\largeâ‡’~y\) = \(\large\frac{my_2~+~ny_1}{m~+~n}\)

So, the coordinates of the point \(M(x,y)\) which divides the line segment joining points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) internally in the ratio \(m:n\) are

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

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 the 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 \(A(x_1, y_1). C(x,y)\) is a point which divides the line \(OA\) in the ratio \(2:1\), where point \(O\) is origin.

\(â‡’~OC:CA\) = \(2:1\)

To find coordinates of the point \(C\) , three lines \(CD, ~AB ~and~ CE \) are drawn such that \(CD\) and \(AB\) are perpendicular to \(x\) – axis and \(CE\) is parallel to \(x\) – axis.

\(âˆ COD\) = \(âˆ ACE\) [Corresponding angles]

\(âˆ CDO\) = \(âˆ AEC\) = \(90Â°\)

By \(AA\) criterion for similarity of two triangles,

\(âˆ†OCD\) ~ \(âˆ†CAE\)

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

\(\frac{OC}{CA}\) = \(\frac{CD}{AE}\) = \(\frac{OD}{CE}\) Â Â Â Â Â Â Â —(1)

Since coordinates of \(C\) are (\(x,y\)),

\(OD\) = \(x\) units

\(BE\) = \(CD\) = \(y\) units

\(CE\) = \(BD\) = \(OB~-~OD\) = (\(x_1~-~x\)) units

\(AE\) = \(AB~-~BE\) = \((y_1~-~y)\) units

Also, \(OC:CA\) = \(2:1\)

\(â‡’~OC\) = \(2a\) units and \(CA\) = \(a\) units, where \(a\) is a constant.

Equation(1) becomes

\(\large \frac{2a}{a}\) = \(\large \frac{x}{x_1~-~x}\) = \(\large \frac{y}{y_1~-~y}\)

\(\large \frac{x}{x_1~-~x}\) = \(2\)

\(\large â‡’~x\) = \(\large 2(x_1~-~x)\)

\(\large â‡’~3x\) = \(\large 2x_1\)

\(\large â‡’~x\) = \(\large \frac{2}{3}~x_1\)

Similarly,

\(\large \frac{y}{y_1~-~y}\) = \(\large 2\)

\(\large â‡’~3y\) = \(\large 2y_1\)

\(\large y\) = \(\large \frac{2}{3}~ y_1\)

Therefore, the coordinates of the point \(C\) is are (\(\large \frac{2}{3}~ x_1, \frac{2}{3}~ y_1\)).

In general, the coordinates of a point which divides the line joining points with coordinates (\(x_1, y_1\)) and (\(x_2, y_2\)) in the ratio \(m:n\) is calculated from here on.

If the point \(M\) divides the line segment joining points \(P\) and \(Q\) internally in the ratio \(k:1\), then coordinates of \(M\) will be

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

What if the point \(M\) which divides the **line segment** joining points \(P(x_1, y_1)\) and \(Q(x_2,y_2)\) is midpoint of line segment \(PQ\)?

If \(M\) is the midpoint, then \(M\) divides the line segment \(PQ\) in the ratio \(1:1\), i.e. \(m\) = \(n\) = \(1\).

Coordinates of point \(M\) are

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

Therefore, **coordinates of a point** which is the midpoint of line segment joining points \((x_1,y_1)\) and \(Q(x_2,y_2)\) are,

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

*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 \(P(x,y)\) be the point which divides the line segment.

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

\(m\) = \(3\), \(n\) = \(2\)

Coordinates of \(P\) are,

\(x\) = \(\large\frac{3~Ã—~-~5~+~2~Ã—~4}{3~+~2}\) Â Â Â Â Â \(y\) = \(\large \frac{3Ã—~-~4~+~2~Ã—~6}{3~+~2}\)

\(â‡’~x\) = \(\large \frac{-15~+~8}{5}\) Â Â Â Â Â \(â‡’~y\) = \(\large \frac{-12~+~12}{5}\)

\(â‡’~x\) = \(\large -\frac{7}{5}\) Â Â Â Â Â Â Â Â Â \(â‡’~y\) = \(0\)

*Example:* The 4 vertices of a parallelogram are \(A\)(-2,3), \(B\)(3,-1), \(C\)(p,q) and \(D\)(-1,9). Find the value of \(p\) and \(q\).

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

Let \(O\) be the point at which diagonals intersect.

Coordinates of mid-points of both \(\overleftrightarrow{AC}\) and \(\overleftrightarrow{BD}\) will be same.

Therefore,

Using midpoint section formula,

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

\(\large \frac{-2~+~p}{2}\) = \(\large \frac{-1+3}{2}\)

\(\large \Rightarrow ~-2~+~p\) = \(2\), \(p\) = \(2~+~2\) = \(4\)

Similarly,

\(\large\frac{3~+~q}{2}\) = \(\frac{9~-~1}{2}\)

\(3~+~q\) = \(8\), \(q\) = \(5\)<

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