# Section Formula & Conic Sections

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 (x1,y1) and (x2,y2) 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.

$∠MPS$ = $∠QMB$ [Corresponding Angles]

$∠MSP$ = $∠QBM$ = $90°$

By $AA$ similarity criterion,

$∆PMS$ ~ $∆MQB$

$⇒~\frac{PM}{MQ}$ = $\frac{PS}{MB}$ = $\frac{MS}{QB}$ = $\frac{m}{n}$ —(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

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

$\large ⇒~\frac{m}{n}$ = $\large \frac{x~-~x_1}{x_2~-~x}$

$\large⇒~x$ = $\large\frac{mx_2~+~nx_1}{m~+~n}$

Similarly,

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

$\large⇒~y$ = $\large\frac{my_2~+~ny_1}{m~+~n}$

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.

$⇒~OC:CA$ = $2:1$

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.

$∠COD$ = $∠ACE$ [Corresponding angles]

$∠CDO$ = $∠AEC$ = $90°$

By $AA$ criterion for similarity of two triangles,

$∆OCD$ ~ $∆CAE$

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

$\frac{OC}{CA}$ = $\frac{CD}{AE}$ = $\frac{OD}{CE}$              —(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, $OC:CA$ = $2:1$

$⇒~OC$ = $2a$ units and $CA$ = $a$ units, where $a$ is a constant.

Equation(1) becomes

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

$\large \frac{x}{x_1~-~x}$ = $2$

$\large ⇒~x$ = $\large 2(x_1~-~x)$

$\large ⇒~3x$ = $\large 2x_1$

$\large ⇒~x$ = $\large \frac{2}{3}~x_1$

Similarly,

$\large \frac{y}{y_1~-~y}$ = $\large 2$

$\large ⇒~3y$ = $\large 2y_1$

$\large y$ = $\large \frac{2}{3}~ y_1$

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

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

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

$x$ = $\large\frac{3~×~-~5~+~2~×~4}{3~+~2}$          $y$ = $\large \frac{3×~-~4~+~2~×~6}{3~+~2}$

$⇒~x$ = $\large \frac{-15~+~8}{5}$           $⇒~y$ = $\large \frac{-12~+~12}{5}$

$⇒~x$ = $\large -\frac{7}{5}$                  $⇒~y$ = $0$

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,

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

$3~+~q$ = $8$, $q$ = $5$<