Advertisements
Advertisements
Question
Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?
Advertisements
Solution
\[∆ = \frac{1}{2}\begin{vmatrix}1 & 4 & 1 \\ 2 & 3 & 1 \\ - 5 & - 3 & 1\end{vmatrix}\]
\[ = \frac{1}{2}\begin{vmatrix}1 & 4 & 1 \\ 1 & - 1 & 0 \\ - 5 & - 3 & 1\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\]
\[ = \frac{1}{2}\begin{vmatrix}1 & 4 & 1 \\ 1 & - 1 & 0 \\ - 6 & - 7 & 0\end{vmatrix} \left[\text{ Applying }R_3 \to R_3 - R_1 \right]\]
\[ = \frac{1}{2}\begin{vmatrix}1 & - 1 \\ - 6 & - 7\end{vmatrix}\]
\[ = \frac{1}{2}\left( - 7 - 6 \right)\]
\[ = \frac{13}{2}\text{ square units }\left[ \because\text{ Area cannot be negative }\right]\]
Therefore, (1, 4), (2, 3) and (−5, −3) are not collinear because,
APPEARS IN
RELATED QUESTIONS
Solve the system of linear equations using the matrix method.
2x + y + z = 1
x – 2y – z = `3/2`
3y – 5z = 9
If A = `[(2,-3,5),(3,2,-4),(1,1,-2)]` find A−1. Using A−1 solve the system of equations:
2x – 3y + 5z = 11
3x + 2y – 4z = –5
x + y – 2z = –3
Evaluate
\[\begin{vmatrix}2 & 3 & - 5 \\ 7 & 1 & - 2 \\ - 3 & 4 & 1\end{vmatrix}\] by two methods.
Find the value of x, if
\[\begin{vmatrix}2 & 3 \\ 4 & 5\end{vmatrix} = \begin{vmatrix}x & 3 \\ 2x & 5\end{vmatrix}\]
Evaluate the following determinant:
\[\begin{vmatrix}67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}6 & - 3 & 2 \\ 2 & - 1 & 2 \\ - 10 & 5 & 2\end{vmatrix}\]
Without expanding, show that the value of the following determinant is zero:
\[\begin{vmatrix}a + b & 2a + b & 3a + b \\ 2a + b & 3a + b & 4a + b \\ 4a + b & 5a + b & 6a + b\end{vmatrix}\]
\[\begin{vmatrix}b + c & a & a \\ b & c + a & b \\ c & c & a + b\end{vmatrix} = 4abc\]
Prove the following identity:
\[\begin{vmatrix}2y & y - z - x & 2y \\ 2z & 2z & z - x - y \\ x - y - z & 2x & 2x\end{vmatrix} = \left( x + y + z \right)^3\]
Using determinants prove that the points (a, b), (a', b') and (a − a', b − b') are collinear if ab' = a'b.
Find values of k, if area of triangle is 4 square units whose vertices are
(k, 0), (4, 0), (0, 2)
Prove that :
\[\begin{vmatrix}\left( b + c \right)^2 & a^2 & bc \\ \left( c + a \right)^2 & b^2 & ca \\ \left( a + b \right)^2 & c^2 & ab\end{vmatrix} = \left( a - b \right) \left( b - c \right) \left( c - a \right) \left( a + b + c \right) \left( a^2 + b^2 + c^2 \right)\]
Prove that :
Prove that :
Prove that
2x − 3y − 4z = 29
− 2x + 5y − z = − 15
3x − y + 5z = − 11
3x + y = 5
− 6x − 2y = 9
Write the value of the determinant
\[\begin{bmatrix}2 & 3 & 4 \\ 2x & 3x & 4x \\ 5 & 6 & 8\end{bmatrix} .\]
Write the cofactor of a12 in the following matrix \[\begin{bmatrix}2 & - 3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & - 7\end{bmatrix} .\]
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.
Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant
The other factor in the value of the determinant is
Solve the following system of equations by matrix method:
3x + 4y − 5 = 0
x − y + 3 = 0
Show that each one of the following systems of linear equation is inconsistent:
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prices at the rate of Rs. x, y and z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total price money of Rs. 37000 and the second institution decided to award respectively 5, 3 and 4 employees with a total price money of Rs. 47000. If all the three prices per person together amount to Rs. 12000 then using matrix method find the value of x, y and z. What values are described in this equations?
Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award ₹x each, ₹y each and ₹z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹2,200. School Q wants to spend ₹3,100 to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as school P). If the total amount of award for one prize on each values is ₹1,200, using matrices, find the award money for each value.
Apart from these three values, suggest one more value which should be considered for award.
2x + 3y − z = 0
x − y − 2z = 0
3x + y + 3z = 0
The system of linear equations:
x + y + z = 2
2x + y − z = 3
3x + 2y + kz = 4 has a unique solution if
Consider the system of equations:
a1x + b1y + c1z = 0
a2x + b2y + c2z = 0
a3x + b3y + c3z = 0,
if \[\begin{vmatrix}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{vmatrix}\]= 0, then the system has
The system of equations:
x + y + z = 5
x + 2y + 3z = 9
x + 3y + λz = µ
has a unique solution, if
(a) λ = 5, µ = 13
(b) λ ≠ 5
(c) λ = 5, µ ≠ 13
(d) µ ≠ 13
Prove that (A–1)′ = (A′)–1, where A is an invertible matrix.
`abs (("a"^2, 2"ab", "b"^2),("b"^2, "a"^2, 2"ab"),(2"ab", "b"^2, "a"^2))` is equal to ____________.
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`
Let A = `[(i, -i),(-i, i)], i = sqrt(-1)`. Then, the system of linear equations `A^8[(x),(y)] = [(8),(64)]` has ______.
