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Section Formula

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Section Formula - Lesson Summary

Let AB be a line segment in space. We have another point C on AB, which divides the line segment into two parts in the ratio m : n.

Finding the coordinates of point C that divides line segment AB:

Let the coordinates of point A be (x1,y1,z1) and point B be (x2,y2,z2).

Let the coordinates of point C be (x,y,z). Perpendiculars from points A, B and C are dropped onto the X - Y plane.

Let the feet of the perpendiculars from points A, C and B on the X - Y plane be P, Q and R, respectively. Lines AP, CR and BQ are parallel to each other. The line joining points P, Q and R lies on the X - Y plane.


Draw a line parallel to line PQ and passing through point C. This line intersects line PA externally. Let the point of intersection be D. Let the point of intersection of this line with BQ be E.

It can be observed that quadrilaterals DPRC and CRQE are parallelograms since lines AP, CR and EQ are parallel to each other, and lines DC, CE, PR and RQ are also parallel to each other.

Since AD and BE are parallel to each other, in triangles ADC and BEC, √ACD = √BCE (Vertically opposite angles), √ADC = √BEC (Alternate angles).
This implies that triangles ADC and BEC are similar

ΔADC ~ ΔBEC

⇒ AC/CB = DA/BE

⇒ m/n = DA/BE

⇒ m/n = DA/BE = DP-AP / BQ - EQ

⇒ m/n = CR-AP / BQ -CR [Since DPRC and CRQE are parallelograms]

Substituting the values, we get

⇒ m/n = z-z1 / z2 - z

⇒ m(z2 - z) = n(z-z1)

    mz2 - mz = nz - nz1

    mz2 + nz1 = nz + mz

     mz2 + nz1 = z(n + m)

(mz2 + nz1)/(n + m) = z

Therefore, the z coordinate of point C that divides line segment AB in the ratio m:n is

z = (mz2 + nz1)/(n + m)

Similarly, using the same method, the other coordinates, x and y, of point C can be found.

Therefore, the coordinates of the point that divides the line segment in the ratio m : n is

C = (((mx2 + nx1)/(m + n)),((my2 + ny1)/(m + n)),((mz2 + nz1)/(m + n)))

Some special cases of section formula

Consider line segment AB in space. If point C divides AB exactly into two parts, then C is dividing AB in the ratio 1:1.

Therefore, m:n = 1:1


This implies that we can consider point C as the midpoint of AB.

Substituting in the standard formula, we get C






C = (((1×x2 + 1 × x1)/(1 + 1)),((1 × y2 + 1 × y1)/(1 + 1)),((1 × z2 + 1 × z1)/(1 + 1)))

Midpoint : ((x1+x2)/2,(y1+y2)/2,(z1+z2)/2).

If point C divides the line segment in the ratio of 1:k, we get

C = (((1×x2 + k × x1)/(1 + k)),((1 × y2 + k × y1)/(1 + k)),((1 × z2 + k × z1)/(1 + k)))

Therefore, the coordinates of the point that divides a line segment in the ratio 1:k is

C = (((kx2 + x1)/(1 + k)),((ky2 + y1)/(1 + k)),((kz2 + z1)/(1 + k))).

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