Class 10

### 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