JEE Exam > JEE Videos > Mathematics (Maths) Class 12 > Examples : Equation of a line
Examples : Equation of a line Video Lecture | Mathematics (Maths) Class 12 - JEE
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FAQs on Examples : Equation of a line Video Lecture - Mathematics (Maths) Class 12 - JEE
| 1. What is the equation of a line? |  |
Ans. The equation of a line is a mathematical representation that relates the x and y coordinates of points on the line. It is typically written in the form y = mx + b, where m represents the slope of the line and b represents the y-intercept.
| 2. How do you find the slope of a line? | |
Ans. The slope of a line can be determined by calculating the ratio of the vertical change (change in y-coordinates) to the horizontal change (change in x-coordinates) between two points on the line. This can be expressed as the formula m = (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are any two points on the line.
| 3. What is the significance of the y-intercept in the equation of a line? | |
Ans. The y-intercept in the equation of a line represents the point where the line intersects the y-axis. It indicates the value of y when x is equal to zero. In the equation y = mx + b, the value of b represents the y-intercept, and it helps determine the starting point of the line on the y-axis.
| 4. Can two lines have the same slope but different y-intercepts? | |
Ans. Yes, two lines can have the same slope but different y-intercepts. The slope only describes the steepness or slant of the line, while the y-intercept determines the vertical position of the line. Even if two lines have the same slope, they can still be shifted up or down along the y-axis, resulting in different y-intercepts.
| 5. How can I graph a line using its equation? | |
Ans. To graph a line using its equation, you can start by identifying the y-intercept from the equation. Plot this point on the y-axis. Then, use the slope of the line to find additional points. The slope tells you how much the line rises or falls as you move horizontally. From the y-intercept, you can move vertically and horizontally according to the slope to plot more points. Finally, connect all the plotted points to form the line.
Text Transcript from Video
hello friends this video on three
dimensions about 2.7 is brought to you
by exam clear calm no more for you from
exam before watching this video please
make sure that you have watched part 1
to part 6 a say for example it says that
we will find the Cartesian equation of
the line which passes through this and
parallel entities so till now what we
have done we have some point here let's
suppose -2 4 and minus 5 is a vector and
this pass this is bad let do some L
vector but in this case instead of
l-vector I know that equation of the
line is of this form X plus 3 by 3 y
minus 4 by 5 Z plus 5 6 and this is some
lambda Y so if you can convert this guy
into my L vector form then I can solve
the question because till now I know
that our vector is nothing but E Plus
lambda into a vector
correct they can convert this guy this
guy in the elevator form my problem
solvers do that please note here we are
interested in L vector we are not
interested in the equation of this line
right this is the equation of the line
from this equation we will find the L
vector so if you know this this equation
is nothing but this is nothing but
equation of line passing through X plus
3 so minus 3 y minus 4 so plus 4 z plus
8 minus 8 so it passes through this and
parallel to some vector L that is 3i v
j6 k 3i plus IJ plus
correct that means this equation this is
the creation of this line L is this that
means this denotes our line equation
which passes through point this and
parallel to some L vector and sell
vector will be 3i plus 5 j plus 6
please note
but I am saying that I have X plus 3 by
3 is equal to Y minus 4 by 5 squared is
n plus 8 by 6 this is nothing but
equation of line that passes through
minus 3 4 & 8 and is parallel to 3i plus
Phi J plus 6 K that is clear now if my
this means that my L vector is 3i plus
Phi J Mascis sorry why see this whole
thing denote some equation of line that
is parallel to this L vector and passes
through this one I am interested only in
L vector so I got this electron from
this equation this is the equation of
line and this is just L rectum and I am
looking only for L vector right so now I
have a I try of l-vector so my equation
of mine is very simple r is good to a
vector plus lambda vector a vector is
given that is minus 2i for J minus PI K
now at this point let us say minus 2i
plus 4 J cap minus phibn plus some
lambda into L vector electrons this one
we found this 3 i cap plus i HD cam plus
60 correct and this is equal to what
minus 2 I plus C lambda so u minus Q
plus 3 lambda I can laughs 4 plus 5
lambda J cap plus plus minus 5 plus 6 9
so this is my equation of line which
passes through point this and parallel
to this the same equation I can write in
this fashion also X plus 2 by 3 is equal
to Y minus 4 by 5 is equal to Z minus
minus 5 is n plus 5 by
how I wrote this see I use the right
method you can use a conventional method
also where you can say that this guy is
X I plus YJ plus ZK all right then you
compare these guys you get lambdas and
you create you get the same answer
actually I'm using a shortcut there I
know that if my am equation of this
fashion X minus x1 by a is equal to Y
minus 1 by B is equal to Z minus z1 by C
this is the equation of the line which
passes through x1 y1 z1 here x1 y1 z1 is
minus 2 plus 4 minus minus 2 plus 4
minus 5 this is minus 2 this is plus 4
and this is right and that is parallel
to a vector a B C and this guy is ABC
here in this case so this guy is 3/5
directly I've used this formula to get
this equation if we don't know this
formula directly if you don't want to
use the formula then you can use the
normal method here I've got this one and
this I can say anything but X I plus YJ
plus ZK because the variable thing right
any any any point in this mine right X Y
Z and you equate these two guys example
you say that minus 2 plus 3 lambda is
equal to X or you say or or you say X
lambda is equal to X plus 2x3
similarly you say 4 plus 5 lambda is
equal to Y or you say lambda is equal to
Y minus 4 by 5 or you say here minus 5
plus 6 lambda is equal to Z or you say
lambda is equal to Z plus 5 by 6 and all
our lambda so you can say that X plus 2
by 3 what is this guy is equal to Y
minus 4 by 5 this guy is going to Z plus
5 s so you got this equation so you can
use any of these you can directly the
formula or you can use the normal
operation of using solutions and that's
how you make the solution so the cashier
will be the given I creation of the line
for this from this equation
the line we found the line vector
please note equation of the line is
different from mine vector so this was
the equation of the line in Cartesian
form so with this we form the line
vector because we are interested only in
the line vector L here right we are not
interested in the equation of this line
duration of the in the line vector of
this line because we are saying that the
final equation which we get what this
line is a vector plus lambda into line
vector all this let us take an example
we have to find the Cartesian equation
of this line the card a ship which is
this we have to find the vector form
this is pretty simple this denotes this
denotes line passing through X minus 5
so this is 5y minus minus 4 so it is
minus 4z minus 6 6 is passing through
this and parallel to 3i plus 7 j + g i7
j mr why because we know that if the
line is of this form X minus x1 by a is
equal to Y minus y1 by P is equal to z2
minus z1 by C that means the line passes
through x1 y1 z1 and it is parallel to a
Z that's how it is right we know this
you compare this guy with this guy we
found at X minus 5 by 1 was minus 4 and
Z 1 is 6 now if this is I know that
there is a line that passes through this
and parent this so this is my you can
say a vector a vector is Phi I minus 4 J
plus 16 and this guy is L vector 3i plus
7 j plus TP same thing I have this is
a vector I have and some L vector I have
so this is going to find equation that
line passes through this a vector and
parallel to this so let assume that this
line is lambda L vector so my equation
of the line will be our victim right
because R can be any point in this line
right so this R vector is nothing but a
vector plus lambda L vector a vector is
this guy Phi I minus 4 J plus 6 K is my
a vector plus lambda into L vector this
guy 3 7 2 3 I plus 7 J plus 2 K right so
with this I get Phi plus 3 lambda I plus
minus both plus m and lambda J plus 6
plus 2 lambda 2 lambda J this is my
equation of the same line in vector form
very easy a difficult thing
so the Cartesian form was given from
this we found our vector form similarly
if my vector form is given I can find
the condition thank you visit exam
viacom to watch free education videos
try free on my debts get the best
quality study materials study from the
best tutors and mentors and much more
thanks and screen
dimensions about 2.7 is brought to you
by exam clear calm no more for you from
exam before watching this video please
make sure that you have watched part 1
to part 6 a say for example it says that
we will find the Cartesian equation of
the line which passes through this and
parallel entities so till now what we
have done we have some point here let's
suppose -2 4 and minus 5 is a vector and
this pass this is bad let do some L
vector but in this case instead of
l-vector I know that equation of the
line is of this form X plus 3 by 3 y
minus 4 by 5 Z plus 5 6 and this is some
lambda Y so if you can convert this guy
into my L vector form then I can solve
the question because till now I know
that our vector is nothing but E Plus
lambda into a vector
correct they can convert this guy this
guy in the elevator form my problem
solvers do that please note here we are
interested in L vector we are not
interested in the equation of this line
right this is the equation of the line
from this equation we will find the L
vector so if you know this this equation
is nothing but this is nothing but
equation of line passing through X plus
3 so minus 3 y minus 4 so plus 4 z plus
8 minus 8 so it passes through this and
parallel to some vector L that is 3i v
j6 k 3i plus IJ plus
correct that means this equation this is
the creation of this line L is this that
means this denotes our line equation
which passes through point this and
parallel to some L vector and sell
vector will be 3i plus 5 j plus 6
please note
but I am saying that I have X plus 3 by
3 is equal to Y minus 4 by 5 squared is
n plus 8 by 6 this is nothing but
equation of line that passes through
minus 3 4 & 8 and is parallel to 3i plus
Phi J plus 6 K that is clear now if my
this means that my L vector is 3i plus
Phi J Mascis sorry why see this whole
thing denote some equation of line that
is parallel to this L vector and passes
through this one I am interested only in
L vector so I got this electron from
this equation this is the equation of
line and this is just L rectum and I am
looking only for L vector right so now I
have a I try of l-vector so my equation
of mine is very simple r is good to a
vector plus lambda vector a vector is
given that is minus 2i for J minus PI K
now at this point let us say minus 2i
plus 4 J cap minus phibn plus some
lambda into L vector electrons this one
we found this 3 i cap plus i HD cam plus
60 correct and this is equal to what
minus 2 I plus C lambda so u minus Q
plus 3 lambda I can laughs 4 plus 5
lambda J cap plus plus minus 5 plus 6 9
so this is my equation of line which
passes through point this and parallel
to this the same equation I can write in
this fashion also X plus 2 by 3 is equal
to Y minus 4 by 5 is equal to Z minus
minus 5 is n plus 5 by
how I wrote this see I use the right
method you can use a conventional method
also where you can say that this guy is
X I plus YJ plus ZK all right then you
compare these guys you get lambdas and
you create you get the same answer
actually I'm using a shortcut there I
know that if my am equation of this
fashion X minus x1 by a is equal to Y
minus 1 by B is equal to Z minus z1 by C
this is the equation of the line which
passes through x1 y1 z1 here x1 y1 z1 is
minus 2 plus 4 minus minus 2 plus 4
minus 5 this is minus 2 this is plus 4
and this is right and that is parallel
to a vector a B C and this guy is ABC
here in this case so this guy is 3/5
directly I've used this formula to get
this equation if we don't know this
formula directly if you don't want to
use the formula then you can use the
normal method here I've got this one and
this I can say anything but X I plus YJ
plus ZK because the variable thing right
any any any point in this mine right X Y
Z and you equate these two guys example
you say that minus 2 plus 3 lambda is
equal to X or you say or or you say X
lambda is equal to X plus 2x3
similarly you say 4 plus 5 lambda is
equal to Y or you say lambda is equal to
Y minus 4 by 5 or you say here minus 5
plus 6 lambda is equal to Z or you say
lambda is equal to Z plus 5 by 6 and all
our lambda so you can say that X plus 2
by 3 what is this guy is equal to Y
minus 4 by 5 this guy is going to Z plus
5 s so you got this equation so you can
use any of these you can directly the
formula or you can use the normal
operation of using solutions and that's
how you make the solution so the cashier
will be the given I creation of the line
for this from this equation
the line we found the line vector
please note equation of the line is
different from mine vector so this was
the equation of the line in Cartesian
form so with this we form the line
vector because we are interested only in
the line vector L here right we are not
interested in the equation of this line
duration of the in the line vector of
this line because we are saying that the
final equation which we get what this
line is a vector plus lambda into line
vector all this let us take an example
we have to find the Cartesian equation
of this line the card a ship which is
this we have to find the vector form
this is pretty simple this denotes this
denotes line passing through X minus 5
so this is 5y minus minus 4 so it is
minus 4z minus 6 6 is passing through
this and parallel to 3i plus 7 j + g i7
j mr why because we know that if the
line is of this form X minus x1 by a is
equal to Y minus y1 by P is equal to z2
minus z1 by C that means the line passes
through x1 y1 z1 and it is parallel to a
Z that's how it is right we know this
you compare this guy with this guy we
found at X minus 5 by 1 was minus 4 and
Z 1 is 6 now if this is I know that
there is a line that passes through this
and parent this so this is my you can
say a vector a vector is Phi I minus 4 J
plus 16 and this guy is L vector 3i plus
7 j plus TP same thing I have this is
a vector I have and some L vector I have
so this is going to find equation that
line passes through this a vector and
parallel to this so let assume that this
line is lambda L vector so my equation
of the line will be our victim right
because R can be any point in this line
right so this R vector is nothing but a
vector plus lambda L vector a vector is
this guy Phi I minus 4 J plus 6 K is my
a vector plus lambda into L vector this
guy 3 7 2 3 I plus 7 J plus 2 K right so
with this I get Phi plus 3 lambda I plus
minus both plus m and lambda J plus 6
plus 2 lambda 2 lambda J this is my
equation of the same line in vector form
very easy a difficult thing
so the Cartesian form was given from
this we found our vector form similarly
if my vector form is given I can find
the condition thank you visit exam
viacom to watch free education videos
try free on my debts get the best
quality study materials study from the
best tutors and mentors and much more
thanks and screen
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About this Video
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Dec 22, 2024 Last updated |
Video Description: Examples : Equation of a line for JEE 2024 is part of Mathematics (Maths) Class 12 preparation.
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