Examples (NCERT) - Straight Lines Video Lecture | Mathematics for Airmen Group X - Airforce X Y / Indian Navy SSR

hello friends this video straight line
parts I went in this brought to you by
example a calm no more fear from exam
before watching this video please make
sure that you have a straight line in
part 1 to part 16 there's a common
problem that says two lines passing
through point in common three intersect
each other at an angle of 60 degree the
slope of one line is to find the
equation of other night let's draw this
problem this is line 1 this is 90 there
are two lines then angle between the
line is 60 degree this is line element
this is language for line L mmm mmm is
good and this point is U comma this is
point P what we have to do we have to
find a equation of line to find the
equation of line and we need either two
points in a new or we need one point and
one slope we know the point that 2 comma
3 is appoint them and do if you can find
the slope of L 2 we can find the
equation now since we know the angle
between these two lines
and we knew slope of this line Delvin
you can find just group of nine head oh
you know this formula that tan theta
that is angle between any two point is
mod of angular to any two line is mod of
M 1 minus M 2 by 1 plus M this is the
equation we have and this value is tan
60 and this is nothing but who correct
or we can say that mod of M is nothing
but 2 minus M 2 by 1 plus 2 m is equal
to if you remove the mod that means this
is either positive or negative you do
move the mod so
we have Jason actually here case 1 is 2
minus M 2 by 1 plus M 2 is equal to 3
and we have case 2 here where we say 2
minus M 2 by 1 plus 2 by M 2 is good -
Wootton please note what we have done
here we have removed the mod so we have
got 2 case 1 is in first case the RFA is
positive in second analysis negative so
we have got two cases one case it is
positive another case of this so it's
all both this equation so in first case
we get 2 minus m2 is equal to root 3
plus 3 M 2 or we get M is equal to minus
root 3 by 1 plus 2 this is first case
here what we get minus M 2 is equal to
minus 1 3 minus 2 root 3 em - all we get
M 2 plus 3 by 1 minus 2 3 so we have got
two values of M V this is 2 minus root 3
by 1 plus 2 root 3 or 2 plus root 3 by 1
minus 2 now we have got two bands I do
so we get two values of the equation of
line into so there can be two possible
condition I can show you how also if L 1
is fixed 60 degree can with this side or
six period with this side also so we can
get to where exactly like this can also
be a line so this can also be a line so
this can be added to and this can be add
to - so they gave it to equation of line
correct
so here also we are getting to the
others now we have two values of slopes
and we have one point 2 comma 3
the equation of the line now here
case one will see u comma three P is 2
by 3 and M is 2 minus 2 3 y 1 plus 2 now
for this the equation is y minus y
naught is equal to M into X minus X so
this becomes y -3 is equal to 2 minus 2
root 3 by 1 plus 2 root 3 into X means
this we can write in this fashion as we
can say Y minus 3 into 1 plus 2 o 3 is
equal to X minus 2 into minus 2 can
solve this equation also we can keep
like further it's all our wish slits all
we'll get this as 2 minus root 3 X this
is the X form minus 1 plus 2 root 3 y is
the Y form class X this is X we're
taking this become the stage ego-
class this becomes minus 2 into 2 minus
rho3 correct again plus 3 this minus
will become plus plus 3 into 1 plus this
is the value so we get 2 minus root 3 X
minus 1 plus 2 3 y plus I will paying
out the value of this this will turn out
to be minus 4 plus 2 3 plus 3 plus 6
squatting it is eight to three I'll find
this two minus three X minus one plus
three y plus plus 6 plus 2 8 3 minus
this is the answer this is the equation
of mine similarly in case two here we
have P is equal to 2 comma 3 and M in
this case was plus root 3 by 1 minus 2
it was a we can use same equation that
is y minus y naught is equal to M into X
minus X naught this becomes or Y minus 3
is equal to 2 plus 3 by 1 minus 2 3 into
X minus or we can write this equation as
y minus 3 into 1 minus 2 is equal to X
minus 2 into plus and if you solve this
we'll get the answer so we have one the
equation of what the
so we have formed the equation motion
says it is the length of perpendicular
from origin to the line whose intercepts
on axis are a and B we have to show the
Hat 1 plus on the B square plus 1 is
equal to 1 by B Square + by C square
this is the XY axis we have this is X
this is y and we are told that B is the
length of perpendicular from origin to
the line whose intercepts on x and y are
a and B there are x intercept is a and y
intercept is B and P is the length of
origin if you take this point we draw a
perpendicular
this led this people now we have to
prove this right this line be length L
so equation of line L this forest line L
the equation will be X by ay plus y by B
is equal 1 because we know both x and y
intercepts so we can write this machine
as X by Y because Y by B squared - 1 now
to draw to find the length of
perpendicular we have to first find the
equation in X plus DX plus dy plus C is
equal to 0 form little signal formula so
we can convert this into single form we
can say that X B plus y minus VB is
equal to 0 this is the equation this is
the equation in X plus dy plus C is
going to 0 for now from 0 0 this is 0 0
from 0 0 you want to find that distance
P so P is nothing but mod of a x1 a is
equal to air B into 0 plus small V sorry
then B we say a into 0 plus C C is minus
ETA by root of a square plus B square
root is here me and V is here this be if
you come where is a this guy is born of
this this is nothing but a B by root of
ay square plus these this is equal to P
Square this equation so what we get a
square B Square by a square plus B
Square now we have to prove 1 by P
Square is equal to 1 by B Square 1 plus
1 by a square so we'll put 1 by B Square
in LHS so 1 by P Square is nothing but a
square plus B square by a square this is
nothing but a square
by a square v square plus v square by a
square v square this is nothing but 1 by
v square plus 1 by and that desirest so
we have proved this what we have done we
knew that X and y-intercept for this
line is a and B so we form the equation
of this line and then from 0 0 we have
to find the distance in that is equal to
B we found a distance that came out to
be a VA who is got the B square that is
equal to B we square it and we found
that this is what we have to thank you
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