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SAI International School
CLASS - X

Mathematics
CHAPTER-7: Co-Ordinate Geom-2Lesson Notes-2

SUBTOPIC : Section Formula

Section Formula

Consider any two points A(x , y ) and B(x , y ) and assume that P (x, y) divides
1
2
1
2
PA m
AB internally in the ratio m : m , i.e.,  1
2
1
PB m 2

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,
PA AQ PQ
PAQ~BPC, So   (1)
BP PC BC

Now, AQ = RS = OS – OR = x – x
1

PC = ST = OT – OS = x – x
2
PQ = PS – QS = PS – AR = y – y
1
BC = BT– CT = BT – PS = y – y
2
m x  x y  y
Substituting these values in (1), we get 1  1  1
m 2 x  x y  y
2
2
m x  x
Taking 1  1 ,  m (x  ) x  m (x  x ) , m x  m x  m x  m x , 
m 2 x  x 1 2 2 1 1 2 1 2 2 1
2
m x  m x
m x  m x  m x  m x , m x  m x  ( m  m x ) ,we have x = 1 2 2 1
2
m  m 2
1
2
1
1
2
2
2
1
2
1
1
1
m y  y m y  m y
and 1  1 , similarly we have y= 1 2 2 1
m 2 y  y m  m 2
1
2
So, the coordinates of the point P(x, y) which divides the line segment joining the points A(x , y ) and
1
1
m x  m x m y  m y
B(x , y ), internally, in the ratio m : m are 1 2 2 1 , 1 2 2 1
2
1
2
2
m  m 2 m  m 2
1
1
This is known as the section formula.
1