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 coordinate 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 coordinates of P are (\(\frac{mx_{2} + nx_{1}}{m + n}\), \(\frac{my_{2} + ny_{1}}{m + n}\)).
`● Distance and Section Formulae
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