Let us, Suppose a telephone company wants to position a relay tower at `P` between `A` and `B` is such a way that the distance of the tower from `B` is twice its distance from `A`. If `P` lies on `AB`, it will divide `AB` in the ratio `1 : 2` (see Fig. 7.9).
If we take `A` as the origin` O`, and `1 km `as one unit on both the axis, the coordinates of `B` will be `(36, 15)`. In order to know the position of the tower, we must know the coordinates of P. Let's find these coordinates
Let the coordinates of `P` be `(x, y)`. Draw perpendiculars from `P` and `B` to the `x`-axis, meeting it in `D` and `E`, respectively. Draw `PC` perpendicular to `BE`. Then, by the `AA` similarity criterion, studied in Chapter 6, `Delta POD` and `Delta BPC` are similar.
Therefore , `(OD)/(PC) = (OP)/(PC) =1/2` , and `(PD)/(BC) = (OP)/(PB) = 1/2`
So, `x/(36-x) = 1/2` and `y/(15- y) = 1/2` .
You can check that` P(12, 5)` meets the condition that `OP : PB = 1 : 2`.
Now let us use the understanding that you may have developed through this example to obtain the general formula.
Consider any two points `A(x_1, y_1)` and `B(x_2, y_2)` and assume that `P (x, y)` divides `AB` internally in the ratio `m_1 : m_2`, i.e.,
`(PA)/(PB) = (m_1)/(m_2)` (see Fig. 7.10).
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 ~ Delta BPC`
Therefore, `(PA)/(BP) = (AQ)/(PC) = (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 (1), we get
`(m_1)/(m_2) = (x- x_1)/( x_2 -x ) = (y -y_1)/( y_2 - y)`
Taking ` (m_1)/(m_2) = (x-x_1)/(x_2 - x) `, we get `x = (m_1 x_2 + m_2 x_1)/( m_1 +m_2)`
Similarly, taking ` (m_1)/(m_2) = (y-y_1)/( y_2 -y)` , we get ` y = ( m_1 y_2 + m_2 y_1)/( m_1 + m_2)`
So, the coordinates of the point P(x, y) which divides the line segment joining the points `A(x_1, y_1)` and `B(x_2, y_2)`, internally, in the ratio `m_1 : m_2` are
` ( (m_1 x_2 + m_2 x_1)/( m_1 + m_2) , ( m_1 y_2 + m_2 y_1)/( m_1 + m_2) )` ............(2)
This is known as the section formula.
This can also be derived by drawing perpendiculars from `A, P` and `B` on the
`y`-axis and proceeding as above.
If the ratio in which `P `divides `AB` is `k : 1`, then the coordinates of the point `P` will be
`( ( k x_2 + x_1)/( k+1) , ( k y_2 + y_1)/(k+1) )`.
Special Case : The mid-point of a line segment divides the line segment in the ratio
`1 : 1`. Therefore, the coordinates of the mid-point `P` of the join of the points `A(x_1, y_1)`
and `B(x_2, y_2)` is
`( (1 * x_1 + 1 * x_2)/( 1+1) , ( 1 * y_1 + 1 * y_2)/( 1+1) ) = ( (x_1 + x_2 )/2, ( y_1 + y_2)/2 )` .
Let us solve a few examples based on the section formula.