# 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

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

Note: If AP : PB = m : n then AP : PB = $$\frac{m}{n}$$ : 1 (since m : n = $$\frac{m}{n}$$ : $$\frac{n}{n}$$. 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 (x$$_{1}$$, y$$_{1}$$) and B (x$$_{2}$$, y$$_{2}$$) internally in the ratio m : n (i.e., $$\frac{AP}{PB}$$ = $$\frac{m}{n}$$) are given by

x = ($$\frac{mx_{2} + nx_{1}}{m + n}$$, y = $$\frac{my_{2} + ny_{1}}{m + n}$$)

Proof:

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

Let A (x$$_{1}$$, y$$_{1}$$) and B (x$$_{2}$$, y$$_{2}$$) 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, $$\frac{AP}{PB}$$ = $$\frac{m}{n}$$)

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

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

AL = y$$_{1}$$, OL = x$$_{1}$$, BM = y$$_{2}$$, OM = x$$_{2}$$, PN = y and ON = x.

By geometry,

AR = LN = ON – OL = (x - x$$_{1}$$);

PS = NM = OM – ON = (x$$_{2}$$ -  x);

PR = PN – RN = PN – AL = (y - y$$_{1}$$)

BS = BM – SM = BM – PN = (y$$_{2}$$ - y)

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

Thus, $$\frac{AP}{PB}$$ = $$\frac{AR}{PS}$$ = $$\frac{PR}{BS}$$

⟹ $$\frac{m}{n}$$ = $$\frac{x - x_{1}}{x_{2} - x}$$ = $$\frac{y - y_{1}}{y_{2} - y}$$

⟹ $$\frac{m}{n}$$ = $$\frac{x - x_{1}}{x_{2} - x}$$ and $$\frac{m}{n}$$ = $$\frac{y - y_{1}}{y_{2} - y}$$

⟹ (m + n)x = (mx$$_{2}$$ + nx$$_{1}$$) and (m + n)y = (my$$_{2}$$ + ny$$_{1}$$)

⟹ x = ($$\frac{mx_{2} + nx_{1}}{m + n}$$ and y = $$\frac{my_{2} + ny_{1}}{m + n}$$)

Therefore, the co-ordinates of P are  ($$\frac{mx_{2} + nx_{1}}{m + n}$$,  $$\frac{my_{2} + ny_{1}}{m + n}$$).

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