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Section Formula

We will proof the definition of section formula.

Section of a Line Segment

Let AB be a line segment joining the points A and B. Let P be any point on the line segment such that AP : PB = λ : 1

Section of a Line Segment

Then, we can say that P divides internally AB is the ratio λ : 1.

Note: If AP : PB = m : n then AP : PB = mn : 1 (since m : n = mn : nn. So, any section by P can be expressed as AP : PB = λ : 1

Definition of section formula: The coordinates (x, y) of a point P divides the line segment joining A (x1, y1) and B (x2, y2) internally in the ratio m : n (i.e., APPB = mn) are given by

x = (mx2+nx1m+n, y = my2+ny1m+n)


Proof:

Let X’OX and YOY’ are the co-ordinate axes.

Let A (x1, y1) and B (x2, y2) be the end points of the given line segment AB.

Let P(x, y) be the point which divides AB in the ratio m : n.

Then, APPB = mn)

We want to find the coordinates (x, y) of P.

Draw AL ⊥ OX; BM ⊥ OX; PN ⊥ OX; AR ⊥ PN; and PS ⊥ BM

AL = y1, OL = x1, BM = y2, OM = x2, PN = y and ON = x.

By geometry,

AR = LN = ON – OL = (x - x1);

PS = NM = OM – ON = (x2 -  x);

PR = PN – RN = PN – AL = (y - y1)

BS = BM – SM = BM – PN = (y2 - y)

Clearly, we see that triangle ARP and triangle PSB are similar and, therefore, their sides are proportional.

Thus, APPB = ARPS = PRBS

mn = xx1x2x = yy1y2y

mn = xx1x2x and mn = yy1y2y

⟹ (m + n)x = (mx2 + nx1) and (m + n)y = (my2 + ny1)

⟹ x = (mx2+nx1m+n and y = my2+ny1m+n)

Therefore, the co-ordinates of P are  (mx2+nx1m+n,  my2+ny1m+n).

 Distance and Section Formulae






10th Grade Math

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