हिंदी

Find the Area of the Triangle with Vertice at the Point: (0, 0), (6, 0) and (4, 3)

Advertisements
Advertisements

प्रश्न

Find the area of the triangle with vertice at the point:

 (0, 0), (6, 0) and (4, 3)

Advertisements

उत्तर

\[∆ = \frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ 6 & 0 & 1 \\ 4 & 3 & 1\end{vmatrix} \] 

\[ ∆ = \frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ 6 & 0 & 0 \\ 4 & 3 & 1\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\] 

\[ ∆ = \frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ 6 & 0 & 0 \\ 4 & 3 & 0\end{vmatrix} \left[\text{ Applying }R_3 \to R_3 - R_1 \right]\] 

\[ ∆ = \frac{1}{2}\begin{vmatrix}6 & 0 \\ 4 & 3\end{vmatrix}\] 

\[ ∆ = \frac{1}{2}\left( 18 - 0 \right)\] 

\[ ∆ = \frac{1}{2}\left( 18 \right) = 9\text{ square units }\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Determinants - Exercise 6.3 [पृष्ठ ७१]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 5 Determinants
Exercise 6.3 | Q 1.4 | पृष्ठ ७१

संबंधित प्रश्न

Solve the system of linear equations using the matrix method.

5x + 2y = 3

3x + 2y = 5


The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method.


If A \[\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\] , then show that |3 A| = 27 |A|.

 

Find the value of x, if
\[\begin{vmatrix}2 & 4 \\ 5 & 1\end{vmatrix} = \begin{vmatrix}2x & 4 \\ 6 & x\end{vmatrix}\]


For what value of x the matrix A is singular? 
\[ A = \begin{bmatrix}1 + x & 7 \\ 3 - x & 8\end{bmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12\end{vmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}\cos\left( x + y \right) & - \sin\left( x + y \right) & \cos2y \\ \sin x & \cos x & \sin y \\ - \cos x & \sin x & - \cos y\end{vmatrix}\]


Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}\sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1\end{vmatrix}, where A, B, C \text{ are the angles of }∆ ABC .\]


Prove that

\[\begin{vmatrix}\frac{a^2 + b^2}{c} & c & c \\ a & \frac{b^2 + c^2}{a} & a \\ b & b & \frac{c^2 + a^2}{b}\end{vmatrix} = 4abc\]


Prove that
\[\begin{vmatrix}- bc & b^2 + bc & c^2 + bc \\ a^2 + ac & - ac & c^2 + ac \\ a^2 + ab & b^2 + ab & - ab\end{vmatrix} = \left( ab + bc + ca \right)^3\]


Prove the following identity:

\[\begin{vmatrix}a + x & y & z \\ x & a + y & z \\ x & y & a + z\end{vmatrix} = a^2 \left( a + x + y + z \right)\]

 


​Solve the following determinant equation:

\[\begin{vmatrix}1 & x & x^2 \\ 1 & a & a^2 \\ 1 & b & b^2\end{vmatrix} = 0, a \neq b\]

 


If the points (a, 0), (0, b) and (1, 1) are collinear, prove that a + b = ab.


5x − 7y + z = 11
6x − 8y − z = 15
3x + 2y − 6z = 7


3x + y = 5
− 6x − 2y = 9


x − y + 3z = 6
x + 3y − 3z = − 4
5x + 3y + 3z = 10


If w is an imaginary cube root of unity, find the value of \[\begin{vmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{vmatrix}\]


If A = [aij] is a 3 × 3 scalar matrix such that a11 = 2, then write the value of |A|.

 

If \[x, y \in \mathbb{R}\], then the determinant 

\[∆ = \begin{vmatrix}\cos x & - \sin x  & 1 \\ \sin x & \cos x & 1 \\ \cos\left( x + y \right) & - \sin\left( x + y \right) & 0\end{vmatrix}\]



There are two values of a which makes the determinant  \[∆ = \begin{vmatrix}1 & - 2 & 5 \\ 2 & a & - 1 \\ 0 & 4 & 2a\end{vmatrix}\]  equal to 86. The sum of these two values is

 


Solve the following system of equations by matrix method:
5x + 7y + 2 = 0
4x + 6y + 3 = 0


Show that the following systems of linear equations is consistent and also find their solutions:
2x + 3y = 5
6x + 9y = 15


Show that the following systems of linear equations is consistent and also find their solutions:
2x + 2y − 2z = 1
4x + 4y − z = 2
6x + 6y + 2z = 3


Show that each one of the following systems of linear equation is inconsistent:
2x + 5y = 7
6x + 15y = 13


Show that each one of the following systems of linear equation is inconsistent:
4x − 2y = 3
6x − 3y = 5


If \[A = \begin{bmatrix}3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1\end{bmatrix}\] , find A−1 and hence solve the following system of equations: 

\[A = \begin{bmatrix}1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5\end{bmatrix}\], find AB. Hence, solve the system of equations: x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

Given \[A = \begin{bmatrix}2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5\end{bmatrix}, B = \begin{bmatrix}1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{bmatrix}\] , find BA and use this to solve the system of equations  y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17


x + y − 6z = 0
x − y + 2z = 0
−3x + y + 2z = 0


The existence of the unique solution of the system of equations:
x + y + z = λ
5x − y + µz = 10
2x + 3y − z = 6
depends on


On her birthday Seema decided to donate some money to children of an orphanage home. If there were 8 children less, everyone would have got ₹ 10 more. However, if there were 16 children more, everyone would have got ₹ 10 less. Using the matrix method, find the number of children and the amount distributed by Seema. What values are reflected by Seema’s decision?


Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.


`abs ((1, "a"^2 + "bc", "a"^3),(1, "b"^2 + "ca", "b"^3),(1, "c"^2 + "ab", "c"^3))`


Let A = `[(1,sin α,1),(-sin α,1,sin α),(-1,-sin α,1)]`, where 0 ≤ α ≤ 2π, then:


A set of linear equations is represented by the matrix equation Ax = b. The necessary condition for the existence of a solution for this system is


The value (s) of m does the system of equations 3x + my = m and 2x – 5y = 20 has a solution satisfying the conditions x > 0, y > 0.


If c < 1 and the system of equations x + y – 1 = 0, 2x – y – c = 0 and – bx+ 3by – c = 0 is consistent, then the possible real values of b are


If a, b, c are non-zeros, then the system of equations (α + a)x + αy + αz = 0, αx + (α + b)y + αz = 0, αx+ αy + (α + c)z = 0 has a non-trivial solution if


Using the matrix method, solve the following system of linear equations:

`2/x + 3/y + 10/z` = 4, `4/x - 6/y + 5/z` = 1, `6/x + 9/y - 20/z` = 2.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×