Section Formula with Examples : Part 1 Video Lecture | Mathematics for Grade 11

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before watching this video please make
sure that you have watched introduction
to 3d geometry part 1 - Part 3 now let
me introduce to a new term called
section formula so here a line is
divided in MN in ratio we have done this
for two-dimensional this is puffy
dimensional so let's suppose they had
two points this is point P and this is
point Q this point is x1 y1 z1 and this
point is x2 y2 z2 and this is the line
here since three dimensional plane and
let's suppose there is a point R which
divides the scenes into M into n ratio
then the coordinates of RS this M x2 +
NX 1 by M plus n my 2 plus n by 1 by M
plus n MZ 2 plus NZ 1 by n this is when
there is a internal division this is the
formula very simple is similar to the
2-dimensional only be a valid z they're
the same formula similarly if you are
dividing this externally if this line
for example this is P this is Q right
and there is a point R here
so our here divide this in M this is M
and this is n so if it divides in M into
n Bishan like this then in that case the
formula is M X 2 minus n X 1 by M minus
n M by 2 minus n by 1 by n minus 1 in MZ
2 minus n that 1 by M minus 1 this is
the formula just instead of positive we
have used negative it is all the same
form the CML is formula will be using
this formula to solve the questions of
their have special case where m is good
to and there is it is dividing in one by
one ratio then X comes out to be X 1
plus X 2 by 2 y as well divide both by 2
by 2 and that comes out to be there will
be plus I 2 by 2 same thing if you put M
and M 1 and 1 in that efficiently the
same thing
similarly let's suppose it is dividing
in case to 1 ratio instead of M by M you
see the M by n
cased one where K is equal to M by n so
in that case the formula comes out to be
K X 2 plus X 1 by 1 plus K ki y 2 plus y
minus K and K is n 2 plus it'll buy one
basket this is the same formula except
instead of M and n we are saying in one
same formula let's take some examples we
have to find the coordinates at the
point which divides a line segment
joining 1 minus 2 3 and 3 4 minus 5 min
to entry let's suppose this is a and
this is correct
this is the point now it divides into 2
n 3 and let's suppose this point is P
correct now if you see for point p PX l
pi define that will be nothing but 2
into 3 plus 3 into 1 by 2 plus 3 correct
there's a boundary here so it would be 2
into 3 plus 3 into 1 by 2 plus 3 because
it was M X 1 plus sorry M X 2 plus M X 1
by M last thing that's what I did M is 2
X 2 is 3 plus n is 3 X 1 is 1 by 2 plus
2 so that comes out to be 9 by correct
now for py py will be 2 into 4 plus 3
into minus 2 by 2 plus ly same thing
here MX 2 and X 1 by M plus n M by 2 n
wyman by M plus n so this comes out to
be 8 minus 6 that is 2 by 5 similarly Z
coordinate for this will be 2 into minus
5 you mean 2 minus 5 plus 3 into this
one that is 3 by 2 plus 3 and that comes
out to be minus 1 by so we have all the
values of
so we can say that P is nothing but 9 by
5 2 by 5 and minus 1 by 5 in case of
internal division in M into n ratio now
let's take external now in case of
external division we see that let's
suppose again this is P dash so P dash X
will be nothing but instead of plus here
it will all be minus same thing instead
of plus we have to put minus so that
becomes 2 into 3 minus 3 into 1 by minus
Li and that comes out to be 6 minus 3
that is 3 by minus 1 that is minus 3
similarly P - why
he also instead of plus will put minus
that is 2 into 4 minus 3 into minus 2 by
2 minus 3 and that comes out to be 8
plus 6 14 by minus 1 so minus 14 and PZ
will come out to be 2 B into minus 5
minus 3 into this is again 3 by 2 minus
3 so this is minus 10 minus 6 minus 9
that is minus 19 by minus 1 that is 9 so
if you see that P dash point that point
is nothing but minus 3 minus 14 and this
is the value of P dash which divide this
line a B in to 3 externally and P is the
point which divide the line a B into 2
by 3 internally that is the hence very
simple just we have used the formula in
order let's take for example in this
example we are told that using section
formula be able to prove that these
three points are collinear so let us
assume this is a this is B and this is C
now to prove that these are collinear
what we can do is we can try to prove
something like that that this is point a
this is point B and using section
formula this is Point C so if C points
divides point a and B that means ABC are
collinear there early C can divide so
let's do that
we try to assume that see points divided
at this point a and B into canes to one
ratio
so we'll get C of X that will come out
to be this will be a this will be B so
it will come out to be 2 into K that is
okay - 4 by K plus 1 this is VC X
correct CY will come out to be 4 K plus
6 by K plus 1 similarly C Zed will come
out to be 6 k plus 0 by K plus 6 k plus
tensor 6 k plus 10 by K plus 1 so we
have got this value now we had told that
let's suppose this is Point C so this
value is 14 this value is 0 in this
values minus T correct so if you can
prove that same value of ki comes for
all this so if you solve this we get
some value of K if you solve this we get
some well of K if you solve this we get
some L of K if you get same value of K
in all this equation that means this is
a valid scenario and then we can say
that C divides point a B so let's solve
this to K minus 4 by K plus 2 this is
equal to 14 or 2 K minus 4 is equal to
14 K plus 28 all we get 12 K is equal to
minus 32 correct
- 32 all we get K is equal to minus 32
by 12 correct and this is 3 32:28 sorry
this is k plus 1 this comes out to be
according to k plus one is 40 and this
is minus 14 minus 4 is 18 minus 18 so K
comes out to be minus 18 by 12 and 6 is
comma that is minus 3 by 2 similarly in
this case second case for k plus 6 by K
plus 1 is equal to 0 or we get K is
equal to minus 6 by 4 that is minus 3 by
2 in third case we get 6 k plus 10 by K
plus 1 is equal to minus 2 is minus 2 so
what we get here is what we get here is
6 K plus 10 is equal to minus 2 K minus
3
correct or we get here or we get 8 K is
equal to minus 12 or K is equal to minus
L by 8 that is minus 3 by so if you see
we are getting say values of K in all
these equations and thus we can say that
this is true thank you
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