# Section Formula

In Coordinate Geometry, two points with specific coordinates are connected by a line in a plane, and a third point is located on the line that connects the two points. We must locate the coordinates of the third point, which divides the line into two halves or sections. By using the section formula we can easily get the coordinates of the third point on the line connecting the two points if we know their coordinates and the ratio of the two sections.

The section formula is used to divide a line into two sections (equal or unequal) by a single point. The section formula gives the coordinates of the point that divides a line segment into two parts of a:b lengths.

There are two sorts of section formulas, depending on the placement of the point that divides them into two parts:

• Internal division formula: This section formula is used when the point divides the line segment internally.

Let M and N be the given two points (x1,y1) and (x2,y2) respectively, and P be the point dividing the line-segment MN internally in the ratio a:b, then the section formula for determining the coordinate of a point P is given by:

P(x,y)=(ax2 + bx1a + b, ay2 +by1a +b)

• External division formula: This section formula is used when the point divides the line segment externally.

Let MN be a line segment where M(x1, y1) and N(x2, y2). Let P(x, y) be the point which divides the line segment in the a:b ratio externally.

P(x,y)=(ax2 – bx1a – b, ay2 -by1a -b)

## Application of Section Formula

The incentre, excentre, and centroid of a triangle can all be calculated using the section formula. It’s even utilized in physics to locate the center of mass (COM) of a rigid system of particles.

## Examples

Example 1: Determine the coordinates of the point which divides the segment joining (-3,6) and (5,7) in the ratio 2 : 3.

Solution: Let P(x,y) be the required point that divides the line segment joining the points M(-3,6) and N(5,7).

Here, x1 = -3, y1 = 6, x2 = 5, y2 = 7, a:b =2:3

By section formula for internal division,

x =ax2 + bx1a + b  and      y = ay2 +by1a +b

Putting the values,  x =  2 ×5 + 3×(-3)5   and     y = 2×7 +3×65

x =10-95       and         y = 14 +185

x = 15            and         y = 325

So, the required coordinates are ( 15, 325).

Example 2: In which ratio y-axis divides the line segment joining the points (-3,5) and (6,4).

Solution: When a point is situated on the y-axis then the coordinates of that point will be  (0,y).

So, let point M(0,y) internally divides the line segment joining the points (-3,5) and (6,4) in the ratio a ∶ b.

Here, we shall use the section formula for the internal division for x – coordinate only because it is zero for point M.

x1 = -3, y1 = 5, x2 = 6, y2 = 4

By section formula for internal division, x =x =ax2 + bx1a + b

0 = a ×6 + b×-3a + b

6a=4b

ab=46=23

Therefore, the ratio is a : b = 2 : 3.

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