Collinearity of 3 points Video Lecture | Mathematics (Maths) Class 11 - Commerce

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1 - Part 3 now we'll discuss a new topic
called collinearity of three point so it
is told that if you have two lines which
has same slope and they pass through a
common point that means we are talking
about one level that is two lines will
coincide for example here we a B is one
line let's suppose NVC is one line and
both are panel that means they have same
slope also we say that B is a common
point so in such case B are
hundred-percent confirm that we are
talking about the same line that is easy
alright if we have two lines to the same
home and a common point then this line
will be same line we're talking about a
chamber that means all these points a b
c m or in here also we can say that if
we have two points a b and c then these
points are collinear if and only if the
slope of line area is equal to slope of
line easy so if we have a scenario where
slope of line ay B is equal to stop
applying BC then these points are
quality we will take an example of
correlative three points PQ R Y in our
line the coordinates of P Q R naught H
comma K x1 comma y1 x2 comma y2 and we
have to prove that H minus x1 into y2
minus y1 is equal to 3 minus Phi 1 into
X 2 minus 1 before shot in this let's
draw the slide
this is the line we have drawn and here
we have three points this is P this is Q
and this is the suppose R so we have
three points P Q R P has H comma K Q is
x1 comma y1 and R is x2 comma Y so we
have three points PQ R now I'm going to
prove that H Y minus X 1 R is this minus
x1 y2 minus y1 y2 minus y1 so if you see
this this is nothing but they're
comparing trying to compare the slopes
so this equation we can also write this
equation whatever we have got is white
for it to - this is U on to this we know
that this equation if u divided by x2
minus x1 both side correct x2 minus x1
so this is cancel so this is y2 minus y1
by X 2 minus X 1 and similarly if you
divide this one by H minus x1 why I am
trying to do is I am just trying to
write this equation in the form of
slopes so what we get is y2 minus y1 by
x2 minus x1 is equal to k minus 5n by H
minus x1 y2 minus y1 is y2 minus y1 this
is y2 is y1 this is nothing but M of Q R
and similarly K minus y 1 K minus y1 by
H minus X this is nothing but M of PQ
unless using the reverse engineering way
so if we can that means if you can start
with this
I went in the reverse direction that
means we have to start with this mqr is
going to empty you and that is true also
we know that since there are three
points they write in a straight line so
that means slope of this P Q will be
equal to slope of Q R because this whole
line has a single slope you take PR you
take the bow line PR PQ or QR each of
these lines have similar slope so let us
try to solve them now we can say that
since since since PQ R are collinear so
we can say that M of PQ is equal to M of
Q our slope of PQ is when to stop option
that is what we have got hungry bus
living wave so slope of PQ is nothing
but H sorry K minus y1 by H minus X this
is above P similarly slope of Q R is
nothing but this is white white minus y1
by x2 minus x1 but it is no cure
now from this equation if you take this
one we will cross multiply but it is K
minus y1 into X 2 minus X 1 is equal to
y2 minus y1 you H minus X 1 and that is
what we were asked to so we have proved
it very simple what we do we so when we
are asked to prove something we follow
the reverse engineering way so from
reverse engineering we found that you
have to start somewhere like this M of Q
R is equal to M of PQ and that's what we
have started we know
is possible history because eqr coronial
we started with this slope M before no
answer if it have started with some
other slope we may not have got the
answer
since you want to find the exact answer
want to prove something it's better to
vote you know reverse me and that's what
we have done here here the cushion is
lines to mix plus y minus 3 is equal to
0 5 X plus K Y minus 3 is equal to 0 & 3
X minus y minus 2 is equal to 0 are
concurrent
we have to find the value of K so they
have concurrent means they meet at a
point these three lines they be at a
point
this is l1 is going to this is them so
equation of Elven is 2x plus 5 minus 3
is equal to 0 and 2 is 5 X plus K Y
minus 3 is equal to 0 and L 3 is 3 X
minus y -2 we have to find the value of
K first let's fight weak weight line L 1
L of G if you equate Ln energy will get
the common point the common point is
clear answer let's equate L&L 3 so we
get let's do it here 2x plus y -3 is
equal to 0 and 3x minus y -2 0 correct
if you add these two equation what you
get is we get 5x minus 5 is going to 0
or X is equal to 1 putting this value of
x Square to 1 in any of the equation
let's suppose we will put in the last
equation so we get 3x minus y -2 is
equal to 0 or 3 into 1 minus y -2 is
equal to 0 or Y is equal to 3 minus 2
that is
therefore X in via 1 comma 1 comma so
this P point is 1 comma 1 we have found
this this is a common point now since
this is a common point line f2 will all
satisfy this one because n 2 also has
this point all this line has a smile so
we will put this value in mind nd so we
get 5x plus KY minus 3 is equal to 0 or
then say 5 into 1 plus K into 1 minus 3
is equal to 0 or we get K is equal to Cu
minus 5 is equal to minus 2 so value of
case minus J very simple what we have
done we had three equations we formed
our graph and we found that there should
be a common point that this common to
all is three lines we solved equation 1
and equation 3 we found a common point
common point turned out to be 1 comma 1
and since line L 2 will also pass
through this 1 comma 1 1 comma 1 will
also satisfy line 1 line at do we have
put 1 comma 1 to this line L 2 and we
got our log K that is minus sister let's
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