Class 12 Exam > Class 12 Videos > Example Area of triangle using Determinant
Example Area of triangle using Determinant Video Lecture - Class 12
FAQs on Example Area of triangle using Determinant Video Lecture - Class 12
| 1. What is the determinant method for finding the area of a triangle? |  |
Ans. The determinant method for finding the area of a triangle involves using the coordinates of its vertices. By forming a 3x3 matrix with the x-coordinates of the vertices in the first column, the y-coordinates in the second column, and 1's in the third column, the determinant of this matrix divided by 2 gives the area of the triangle.
| 2. How does the determinant method work in finding the area of a triangle? | |
Ans. The determinant method works by calculating the determinant of a matrix created with the coordinates of the triangle's vertices. This determinant represents the signed area of the parallelogram formed by the vectors connecting the vertices. Taking half of this determinant gives the area of the triangle.
| 3. Can the determinant method be used for any type of triangle? | |
Ans. Yes, the determinant method can be used for any type of triangle, whether it is acute, obtuse, or right-angled. As long as the coordinates of the vertices are known, the determinant method can accurately calculate the area of the triangle.
| 4. Is the determinant method the only way to find the area of a triangle? | |
Ans. No, the determinant method is not the only way to find the area of a triangle. There are other methods available, such as using the lengths of the sides and trigonometry (Heron's formula) or using the base and height of the triangle. However, the determinant method is particularly useful when the coordinates of the vertices are known.
| 5. Are there any limitations or drawbacks to using the determinant method for finding the area of a triangle? | |
Ans. One limitation of the determinant method is that it requires the knowledge of the coordinates of the triangle's vertices. If these coordinates are not available or difficult to obtain, alternative methods may be more suitable. Additionally, the determinant method may become more complex to apply for triangles in higher-dimensional spaces.
Text Transcript from Video
hello friends this video on determining
part 20 is brought you by exam feel calm
no more fear from exam before watching
this video please make sure that you
have watched part 1 to part 19 let's
take one more example we have to prove
that the point a B and C are collinear
so when you say point a b c are
collinear it means the area of triangle
is 0 correct three points are collinear
that means they all quinn sites that
means the triangle form using these
three points the area of the triangle is
zero so what we can do to prove that
these points are collinear i can have to
prove that area of triangle ABC is 0 and
what is the value of area of triangle 1
by 2 determinant of x 1 y 1 1 x 2 y 2 1
and X 3 y 3 1 this guy is X 1 this guy
is y 1 this guy is x2 y2 x3 and y3 so
let's put the raloo which is nothing but
X 1 is a y1 is B plus C and then you
have 1 X 2 is B y 2 is C plus a and 1 X
3 is C and this guy is a plus B by 3 and
then you have 1 so this is my area of
rank now if you can see without even
solving I can find answer how just by
using the property if you see this is a
this is B plus C BC plus 8 see it loves
me if I add these two this becomes a
plus B AC it becomes a plus B plus C and
this also becomes a plus B plus C and I
see that this column and this column
would be proportional there so let's try
this and say r1 is equal to r1 plus r2
so this becomes 1 by 2 R 1 becomes sorry
C 1 is equal to C 1 plus C 2
I am doing column operation right right
again C 1 is equal to C 1 plus C 2 so
this becomes a plus B plus C this
becomes B plus C plus a and this becomes
C plus a plus B correct C plus a plus B
this column is untouched and the next
column is also untouched in this if I
take a plus B plus C common for I get is
1 1 1 B plus C plus a it was B 1 1 1 and
this I have done by saying C 1 is equal
to C 1 why he loves me plus C and as I
can see that two columns are same so I
can see that area is since the area is 0
that means the points are coronium very
simple the concept was for 3 points to
be collinear the area of the triangle
form using D was three two points will
be 0 here you define the value of the K
by area of the triangle is 4 squared and
the vertices are K 0 4 0 0 2 there is
some same portion almost so this is guys
x1 y1 x2 y2 and x3 y3 what is of the
triangle and get over that area is 4
square unit and what is the value of
area 1 by 2 X 1 Y 1 1 x2 y2 1 x3 y3 1 so
he is the area we know replacing the
value of I get X 1 is what K y 1 is 0
and this is 1 here also X 2 is 4 y 2 is
0 and this is 1 X 3 is 0 Y 2 is 2 and
this guy is 1 so if you find this value
if you solve this this comes out nothing
but Y
by 2 into K into 0 into 1 is 0 minus 2
into 1 that is K into minus 2 plus 0
plus 1 into 4 into 2 8 minus 0 into 1
into K this is nothing but 1 by 2 into 8
minus 2 K now I am told that area is 4
so that means 1 by 2 into 8 minus 2 K
can be plus minus 4 because the area is
4 this value can come out to be negative
also if this is negative also minus 4
also in that case also area is 4 because
we take absolute value if it comes out
to be 4 then also building 4 as the
value because we dig absolute value so
in that case 8 minus 2 K is nothing but
plus minus 8 24 so 2 option 1 has 8
minus 2 K is 8 or 8 minus 2 K is minus 8
so in this first case 8 8 cancel I get K
is equal to 0 in the second case I get 2
K is equal to 8 plus 8 is equal to 16 or
I get K is equal to a so I have two
values of K K is equal to 0 or K is
equal to 8 that satisfies this condition
correct so the question asked me when I
find the value of K so there are 2
values okay so the learning from this
example is that when I'm trying to find
the value of change such case please
make sure that you take plus minus area
value don't only take class values
because even it comes out to me because
the area is absolute value area is
absolute value of the determinant by 2
correct
so in that case we'll take plus minus
area now the question is we have to find
the equation of line joining 1 by 2 1
comma 2 and 3 comma 6 using determinant
how can we do this using matrices we
have done we have done in this fashion I
think 1 2 3 6 and then we have XY kind
Vav
like this now we have to do using
determinant how can we do now one thing
to note here is it says there's a line 1
comma 2 is this point 3 comma 6 is this
point and there is any point X comma Y
we have to write the equation for this
line one thing you notice is we take
these 3 points 1 2 X Y and 3 6 the area
of triangle that we get is 0 correct
using these 3 point the area we get is 0
so what we can do is we can form the
area of the triangle to be 0 using these
points and the equation we can get will
be the equation the line I'll tell you
area is nothing but 1 by 2 X 1 Y 1 1 X 2
y 2 1 X 3 y 3 1 correct and this value
is 0 because area of triangle form using
these three points will be 0 because
they all lined in a same line what is
the value of x 1 1 X 2 2 and then 1 so
this is y 2 X 2 is 3 y 2 is 6 and then 1
X 3 is x and y 3 is y and this one now
for this guy the area of triangle is 0
so let's find the value on the left hand
side this is nothing but 1 by 2 into
what this will be doesn't make sense
because this is anyway into 0 so gone so
I can remove this guy so this is nothing
but 1 into 6 minus y -2 into 3 minus X
plus 1 into 3 y minus 6 X correct and
this is equal to 0 if you solve this
what you get you get 6 minus y minus 6
plus 2x
6y minus 12 X is equal to 0 minus 6 plus
6 gone so what I get here is 6 minus 6 y
12 minus 12 X and this is 2 X so this
becomes minus 10 X right so I get minus
10 X plus y minus 5 y plus 5y is equal
to 0 or I get Y is equal to 2 and that
does the equation of my line so what I
have done here I knew two points I just
took the third point x y and then I have
three points and I knew that these three
points are part of the line so the area
of triangle formed using these three
points will be 0 and thus I was able to
get an equation under that so in future
if you get such questions where you're
asked to find the area of or the
equation of the line using determinants
where you have two points given you can
very well do this
thank you visit exam viacom to watch
free educational videos try free online
tests get the best quality study
materials study from the best tutors and
mentors and much more thanks again
part 20 is brought you by exam feel calm
no more fear from exam before watching
this video please make sure that you
have watched part 1 to part 19 let's
take one more example we have to prove
that the point a B and C are collinear
so when you say point a b c are
collinear it means the area of triangle
is 0 correct three points are collinear
that means they all quinn sites that
means the triangle form using these
three points the area of the triangle is
zero so what we can do to prove that
these points are collinear i can have to
prove that area of triangle ABC is 0 and
what is the value of area of triangle 1
by 2 determinant of x 1 y 1 1 x 2 y 2 1
and X 3 y 3 1 this guy is X 1 this guy
is y 1 this guy is x2 y2 x3 and y3 so
let's put the raloo which is nothing but
X 1 is a y1 is B plus C and then you
have 1 X 2 is B y 2 is C plus a and 1 X
3 is C and this guy is a plus B by 3 and
then you have 1 so this is my area of
rank now if you can see without even
solving I can find answer how just by
using the property if you see this is a
this is B plus C BC plus 8 see it loves
me if I add these two this becomes a
plus B AC it becomes a plus B plus C and
this also becomes a plus B plus C and I
see that this column and this column
would be proportional there so let's try
this and say r1 is equal to r1 plus r2
so this becomes 1 by 2 R 1 becomes sorry
C 1 is equal to C 1 plus C 2
I am doing column operation right right
again C 1 is equal to C 1 plus C 2 so
this becomes a plus B plus C this
becomes B plus C plus a and this becomes
C plus a plus B correct C plus a plus B
this column is untouched and the next
column is also untouched in this if I
take a plus B plus C common for I get is
1 1 1 B plus C plus a it was B 1 1 1 and
this I have done by saying C 1 is equal
to C 1 why he loves me plus C and as I
can see that two columns are same so I
can see that area is since the area is 0
that means the points are coronium very
simple the concept was for 3 points to
be collinear the area of the triangle
form using D was three two points will
be 0 here you define the value of the K
by area of the triangle is 4 squared and
the vertices are K 0 4 0 0 2 there is
some same portion almost so this is guys
x1 y1 x2 y2 and x3 y3 what is of the
triangle and get over that area is 4
square unit and what is the value of
area 1 by 2 X 1 Y 1 1 x2 y2 1 x3 y3 1 so
he is the area we know replacing the
value of I get X 1 is what K y 1 is 0
and this is 1 here also X 2 is 4 y 2 is
0 and this is 1 X 3 is 0 Y 2 is 2 and
this guy is 1 so if you find this value
if you solve this this comes out nothing
but Y
by 2 into K into 0 into 1 is 0 minus 2
into 1 that is K into minus 2 plus 0
plus 1 into 4 into 2 8 minus 0 into 1
into K this is nothing but 1 by 2 into 8
minus 2 K now I am told that area is 4
so that means 1 by 2 into 8 minus 2 K
can be plus minus 4 because the area is
4 this value can come out to be negative
also if this is negative also minus 4
also in that case also area is 4 because
we take absolute value if it comes out
to be 4 then also building 4 as the
value because we dig absolute value so
in that case 8 minus 2 K is nothing but
plus minus 8 24 so 2 option 1 has 8
minus 2 K is 8 or 8 minus 2 K is minus 8
so in this first case 8 8 cancel I get K
is equal to 0 in the second case I get 2
K is equal to 8 plus 8 is equal to 16 or
I get K is equal to a so I have two
values of K K is equal to 0 or K is
equal to 8 that satisfies this condition
correct so the question asked me when I
find the value of K so there are 2
values okay so the learning from this
example is that when I'm trying to find
the value of change such case please
make sure that you take plus minus area
value don't only take class values
because even it comes out to me because
the area is absolute value area is
absolute value of the determinant by 2
correct
so in that case we'll take plus minus
area now the question is we have to find
the equation of line joining 1 by 2 1
comma 2 and 3 comma 6 using determinant
how can we do this using matrices we
have done we have done in this fashion I
think 1 2 3 6 and then we have XY kind
Vav
like this now we have to do using
determinant how can we do now one thing
to note here is it says there's a line 1
comma 2 is this point 3 comma 6 is this
point and there is any point X comma Y
we have to write the equation for this
line one thing you notice is we take
these 3 points 1 2 X Y and 3 6 the area
of triangle that we get is 0 correct
using these 3 point the area we get is 0
so what we can do is we can form the
area of the triangle to be 0 using these
points and the equation we can get will
be the equation the line I'll tell you
area is nothing but 1 by 2 X 1 Y 1 1 X 2
y 2 1 X 3 y 3 1 correct and this value
is 0 because area of triangle form using
these three points will be 0 because
they all lined in a same line what is
the value of x 1 1 X 2 2 and then 1 so
this is y 2 X 2 is 3 y 2 is 6 and then 1
X 3 is x and y 3 is y and this one now
for this guy the area of triangle is 0
so let's find the value on the left hand
side this is nothing but 1 by 2 into
what this will be doesn't make sense
because this is anyway into 0 so gone so
I can remove this guy so this is nothing
but 1 into 6 minus y -2 into 3 minus X
plus 1 into 3 y minus 6 X correct and
this is equal to 0 if you solve this
what you get you get 6 minus y minus 6
plus 2x
6y minus 12 X is equal to 0 minus 6 plus
6 gone so what I get here is 6 minus 6 y
12 minus 12 X and this is 2 X so this
becomes minus 10 X right so I get minus
10 X plus y minus 5 y plus 5y is equal
to 0 or I get Y is equal to 2 and that
does the equation of my line so what I
have done here I knew two points I just
took the third point x y and then I have
three points and I knew that these three
points are part of the line so the area
of triangle formed using these three
points will be 0 and thus I was able to
get an equation under that so in future
if you get such questions where you're
asked to find the area of or the
equation of the line using determinants
where you have two points given you can
very well do this
thank you visit exam viacom to watch
free educational videos try free online
tests get the best quality study
materials study from the best tutors and
mentors and much more thanks again
About this Video
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Sep 08, 2025
Last updated
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