Section Formula



A point C(x,y) is present between A(x1,y1) and B(x2,y2). C divides AB internally in the ratio m:n. Then the coordinate of C is given by

C≡( [mx2+nx1)/(m+n)] , [(my2+ny1)/(m+n)] )

Example:
C divides AB internally in the ratio 1:2. A(3,5) and B(1,2). Find C.
Let the coordinate of C be (x,y). Then
Here m=1, n=2, x1= 3, x2 = 1, y1 = 5, y2 = 2.
x = (1×1 + 2×3)/(2+1)
= 7/3
  y= (1×2 + 2×5)/(2+1)
= 12/3
= 4.
Hence the coordinate of C is (7/3,4).
If C divides AB externally in the ratio m:n. Then the coordinate of C is given by

C≡( [mx2− nx1)/(m − n)] , [(my2− ny1)/(m − n)] )


Example:
C divides AB externally in the ratio 2:1. A(3,5) and B(1,2). Find C.
Let the coordinate of C be (x,y). Then
Here m=2, n=1, x1= 3, x2 = 1, y1 = 5, y2 = 2.
x = (2×1 − 1×3)/(2 − 1)
= −1
  y = (2×2 − 1×5)/(2 − 1)
= −1
Hence the coordinate of C is (−1,−1).


Section formula in three dimensional coordinate system

A point C(x,y,z) is present between A(x1,y1,z1) and B(x2,y2,z2). C divides AB internally in the ratio m:n. Then the coordinate of C is given by

C≡( [mx2+nx1)/(m+n)] , [(my2+ny1)/(m+n)] , [(mz2+nz1)/(m+n)] )

Similarly we can find the coordinates of the point when it divides externally as done in two dimensional.


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