Advertisements
Advertisements
Question
Find values of k, if area of triangle is 4 square units whose vertices are
(k, 0), (4, 0), (0, 2)
Advertisements
Solution
\[\text{ If the area of a triangle with vertices (k, 0), (4, 0) and (0, 2) is 4 square units, then }\]
\[\Delta = \frac{1}{2}\begin{vmatrix} k & 0 & 1\\4 & 0 & 1\\0 & 2 & 1 \end{vmatrix} \]
\[ = \frac{1}{2} \left\{ \left( 2 \right) \times \begin{vmatrix} k & 1\\4 & 1 \end{vmatrix} \right\} \left[\text{ Expanding along }C_2 \right]\]
\[ = \left( k - 4 \right)\]
Since area is always +ve, we take its absolute value, which is given as 4 square units .
\[ \Rightarrow ( k - 4 ) = \pm 4\]
\[ \Rightarrow (k - 4) = 4 or (k - 4 ) = - 4\]
\[ \Rightarrow k - 4 = 4 or k - 4 = - 4\]
\[ \Rightarrow k = 8\text{ or }k = 0\]
\[ \Rightarrow k = 8, 0\]
APPEARS IN
RELATED QUESTIONS
Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3
Examine the consistency of the system of equations.
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
Solve the system of linear equations using the matrix method.
5x + 2y = 4
7x + 3y = 5
Solve the system of linear equations using the matrix method.
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
Find the value of x, if
\[\begin{vmatrix}2x & 5 \\ 8 & x\end{vmatrix} = \begin{vmatrix}6 & 5 \\ 8 & 3\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}\sin\alpha & \cos\alpha & \cos(\alpha + \delta) \\ \sin\beta & \cos\beta & \cos(\beta + \delta) \\ \sin\gamma & \cos\gamma & \cos(\gamma + \delta)\end{vmatrix}\]
Evaluate :
\[\begin{vmatrix}x + \lambda & x & x \\ x & x + \lambda & x \\ x & x & x + \lambda\end{vmatrix}\]
\[\begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix} = 4abc\]
\[\begin{vmatrix}- a \left( b^2 + c^2 - a^2 \right) & 2 b^3 & 2 c^3 \\ 2 a^3 & - b \left( c^2 + a^2 - b^2 \right) & 2 c^3 \\ 2 a^3 & 2 b^3 & - c \left( a^2 + b^2 - c^2 \right)\end{vmatrix} = abc \left( a^2 + b^2 + c^2 \right)^3\]
Prove the following identity:
`|(a^3,2,a),(b^3,2,b),(c^3,2,c)| = 2(a-b) (b-c) (c-a) (a+b+c)`
Show that
Find the area of the triangle with vertice at the point:
(2, 7), (1, 1) and (10, 8)
Using determinants show that the following points are collinear:
(5, 5), (−5, 1) and (10, 7)
Find the value of x if the area of ∆ is 35 square cms with vertices (x, 4), (2, −6) and (5, 4).
Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?
Using determinants, find the equation of the line joining the points
(1, 2) and (3, 6)
2x − y = 1
7x − 2y = −7
Prove that
5x + 7y = − 2
4x + 6y = − 3
3x + y + z = 2
2x − 4y + 3z = − 1
4x + y − 3z = − 11
x − y + 3z = 6
x + 3y − 3z = − 4
5x + 3y + 3z = 10
Find the value of the determinant
\[\begin{bmatrix}4200 & 4201 \\ 4205 & 4203\end{bmatrix}\]
If \[A = \begin{bmatrix}0 & i \\ i & 1\end{bmatrix}\text{ and }B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\] , find the value of |A| + |B|.
If \[A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\text{ and }B = \begin{bmatrix}1 & 0 \\ - 1 & 0\end{bmatrix}\] , find |AB|.
Find the maximum value of \[\begin{vmatrix}1 & 1 & 1 \\ 1 & 1 + \sin \theta & 1 \\ 1 & 1 & 1 + \cos \theta\end{vmatrix}\]
If \[\begin{vmatrix}x & \sin \theta & \cos \theta \\ - \sin \theta & - x & 1 \\ \cos \theta & 1 & x\end{vmatrix} = 8\] , write the value of x.
If a > 0 and discriminant of ax2 + 2bx + c is negative, then
\[∆ = \begin{vmatrix}a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0\end{vmatrix} is\]
Let \[A = \begin{bmatrix}1 & \sin \theta & 1 \\ - \sin \theta & 1 & \sin \theta \\ - 1 & - \sin \theta & 1\end{bmatrix},\text{ where 0 }\leq \theta \leq 2\pi . \text{ Then,}\]
Solve the following system of equations by matrix method:
5x + 7y + 2 = 0
4x + 6y + 3 = 0
Solve the following system of equations by matrix method:
3x + 4y − 5 = 0
x − y + 3 = 0
If `alpha, beta, gamma` are in A.P., then `abs (("x" - 3, "x" - 4, "x" - alpha),("x" - 2, "x" - 3, "x" - beta),("x" - 1, "x" - 2, "x" - gamma)) =` ____________.
The value of λ, such that the following system of equations has no solution, is
`2x - y - 2z = - 5`
`x - 2y + z = 2`
`x + y + lambdaz = 3`
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 `|(x + 1, x + 2, x + a),(x + 2, x + 3, x + b),(x + 3, x + 4, x + c)|` = 0, then a, b, care in
If the following equations
x + y – 3 = 0
(1 + λ)x + (2 + λ)y – 8 = 0
x – (1 + λ)y + (2 + λ) = 0
are consistent then the value of λ can be ______.
