derivation of Section formula
Answers
Step-by-step explanation:
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 =
m
n
: 1 (since m : n =
m
n
:
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 (x1, y1) and B (x2, y2) internally in the ratio m : n (i.e.,
AP
PB
=
m
n
) are given by
x = (
mx2+nx1
m+n
, y =
my2+ny1
m+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,
AP
PB
=
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 = 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,
AP
PB
=
AR
PS
=
PR
BS
⟹
m
n
=
x−x1
x2−x
=
y−y1
y2−y
⟹
m
n
=
x−x1
x2−x
and
m
n
=
y−y1
y2−y
⟹ (m + n)x = (mx2 + nx1) and (m + n)y = (my2 + ny1)
⟹ x = (
mx2+nx1
m+n
and y =
my2+ny1
m+n
)
Therefore, the co-ordinates of P are (
mx2+nx1
m+n
,
my2+ny1
m+n
).