Class 10
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class-10 chaper-6

Topic Covered

♦ Section Formula

Section Formula

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.


Q 3109580418

Find the coordinates of the point which divides the line segment joining
the points `(4, – 3)` and `(8, 5)` in the ratio `3 : 1 `internally.
Class 10 Chapter 7 Example 6
Solution:

Let` P(x, y)` be the required point. Using the section formula, we get

`x= ( 3(8) + 1(4) )/(3+1) =7 , y = ( 3(5) +1(-3) )/( 3+1) =3`

Therefore,` (7, 3)` is the required point.
Q 3129680511

In what ratio does the point` (– 4, 6)` divide the line segment joining the
points `A(– 6, 10)` and` B(3, – 8)` ?
Class 10 Chapter 7 Example 7
Solution:

Let` (– 4, 6)` divide AB internally in the ratio `m_1 : m_2`. Using the section
formula, we get

`(-4, 6) = ( (3 m_1 - 6 m_2)/( m_1 + m_2 ) , ( -8 m_1 +10 m_2)/( m_1 + m_2) )` ...................(1)

Recall that if` (x, y) = (a, b)` then `x = a` and `y = b`.

So, ` -4 = (3 m_1 -6 m_2)/( m_1 + m_2) ` and ` 6 = ( -8 m_1 +10 m_2)/( m_1+ m_2)`

Now, ` -4 = (3 m_1 -6 m_2)/( m_1 + m_2) ` gives us

` – 4m_1 – 4m_2 = 3m_1 – 6m_2`

i.e., `7m_1 = 2m_2`

i.e.,` m_1 : m_2 = 2 : 7`

You should verify that the ratio satisfies the y-coordinate also.

Now, ` (-8 m_1 +10 m_2)/(m_1 + m_2) = ( -8 (m_1)/(m_2) +10 )/( (m_1)/(m_2) +1)` (Dividing throughout by `m_2`)

` = (-8 xx 2/7 +10 )/( 2/7 +1) =6`

Therefore, the point `(– 4, 6)` divides the line segment joining the points `A(– 6, 10)` and
`B(3, – 8) `in the ratio `2 : 7`.

Alternatively : The ratio `m_1 : m_2` can also be written as `(m_1)/(m_2) :1` , or ` k :1` . Let `(-4 ,6 )`

divide `AB` internally in the ratio `k : 1.` Using the section formula, we get

` (-4 ,6) = ( (3k -6)/( k+1) , ( -8k +10)/(k+1) )` ...........(2)

So, `-4 = (3k -6 )/(k+1)`

i.e., `– 4k – 4 = 3k – 6`
i.e., `7k = 2`
i.e., `k : 1 = 2 : 7`
You can check for the y-coordinate also.
So, the point` (– 4, 6)` divides the line segment joining the points `A(– 6, 10) `and
`B(3, – 8)` in the ratio `2 : 7`.
Note : You can also find this ratio by calculating the distances `PA` and `PB` and taking
their ratios provided you know that `A, P` and `B `are collinear.
Q 3149680513

Find the coordinates of the points of trisection (i.e., points dividing in
three equal parts) of the line segment joining the points` A(2, – 2)` and `B(– 7, 4)`.
Class 10 Chapter 8 Example 8
Solution:

Let `P` and `Q` be the points of

trisection of `AB` i.e.,` AP = PQ = QB`
(see Fig. 7.11).

Therefore, `P` divides `AB` internally in the ratio `1 : 2`. Therefore, the coordinates of `P`, by
applying the section formula, are

`( (1 (-7) +2 (2) )/( 1+2) , ( 1(4) +2(-2) )/(1+2) ) ` , i.e., `(-1, 0 )`

Now, `Q `also divides AB internally in the ratio `2 : 1`. So, the coordinates of `Q` are

`( ( 2 (-7) + 1(2) )/( 2+1) , ( 2(4) + 1 ( -2) )/( 2+1) )` i.e, `( -4,2 )`



Therefore, the coordinates of the points of trisection of the line segment joining `A` and
`B` are` (–1, 0)` and `(– 4, 2)`.

Note : We could also have obtained `Q` by noting that it is the mid-point of `PB`. So, we
could have obtained its coordinates using the mid-point formula.
Q 3189680517

Find the ratio in which the y-axis divides the line segment joining the
points `(5, – 6)` and` (–1, – 4)`. Also find the point of intersection.
Class 10 Chapter 7 Example 9
Solution:

Let the ratio be `k : 1.` Then by the section formula, the coordinates of the point which divides `AB` in the ratio`

k : 1` are ` ( ( -k +5)/(k+1 ) , (-4k -6)/(k+1)).`

This point lies on the y-axis, and we know that on the y-axis the abscissa is 0.

Therefore, ` (-k+5)/(k+1) = 0`

So, ` k =5`

That is, the ratio is` 5 : 1`. Putting the value of `k = 5`, we get the point of intersection as
`(0, (-13)/3)` .
Q 3119680519

If the points` A(6, 1), B(8, 2), C(9, 4)` and `D(p, 3) `are the vertices of a
parallelogram, taken in order, find the value of p.
Class 10 Chapter 7 Example 10
Solution:

We know that diagonals of a parallelogram bisect each other.
So, the coordinates of the mid-point of `AC =` coordinates of the mid-point of` BD`

i.e., ` ( (6+9)/2 , (1+4)/2 ) = ( (8+p)/2 , (2+3)/2)`

i.e., ` (15/2, 5/2) =( (8+p)/2 , 5/2 )`

so, ` 15/2 = (8+p)/2`

i.e., ` p =7`
 
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