Class 12 Exam > Class 12 Videos > Example Area of Triangle using Determinant - Determinants, Maths, Class 12
Example Area of Triangle using Determinant - Determinants, Maths, Class 12 Video Lecture
FAQs on Example Area of Triangle using Determinant - Determinants, Maths, Class 12 Video Lecture
| 1. What is a determinant in mathematics? |  |
Ans. In mathematics, a determinant is a scalar value that is calculated from the elements of a square matrix. It provides important information about the matrix, such as whether it is invertible or singular, and can be used to solve systems of linear equations.
| 2. How is the determinant used to find the area of a triangle? | |
Ans. The determinant can be used to find the area of a triangle by considering the coordinates of its vertices. By forming a matrix using the x-coordinates and y-coordinates of the vertices, and calculating the determinant of that matrix, the absolute value of the determinant will give the area of the triangle.
| 3. Can the determinant be negative when finding the area of a triangle? | |
Ans. No, the determinant cannot be negative when finding the area of a triangle. Since the area is a measure of the magnitude of a two-dimensional shape, it is always positive or zero. Therefore, when using the determinant method to find the area of a triangle, the absolute value of the determinant is taken to ensure a positive area.
| 4. Are determinants only used to find the area of triangles? | |
Ans. No, determinants have various applications in mathematics and beyond. While they can be used to find the area of triangles, they are also used in linear algebra to solve systems of linear equations, determine whether a matrix is invertible, calculate the volume of parallelepipeds in 3D space, and more.
| 5. What other methods can be used to find the area of a triangle? | |
Ans. Apart from using determinants, there are other methods to find the area of a triangle. One common method is using the formula A = 1/2 * base * height, where the base is the length of the triangle's base and the height is the perpendicular distance from the base to the opposite vertex. Additionally, the Heron's formula can be used when the lengths of all three sides of the triangle are known.
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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Oct 15, 2025
Last updated
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