I. Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the points A, B, and P are collinear.
Therefore, the area of triangle ABP will be zero.
\(\frac{1}{2}\)\(\begin{vmatrix}1&2&1\\3&6&1\\x&y&1\end{vmatrix}\)=0
\(\Rightarrow\)\(\frac{1}{2}\)[1(6-y)-2(3-x)+1(3y-6x)]
=6-y-6+2x+3y-6x=0
\(\Rightarrow\) y=2x
Hence, the equation of the line joining the given points is y = 2x.
II. Let P (x, y) be any point on the line joining points A (3, 1) and B (9, 3). Then, the points A, B, and P are collinear.
Therefore, the area of triangle ABP will be zero.
\(\frac{1}{2}\)\(\begin{vmatrix}3&1&1\\9&3&1\\x&y&1\end{vmatrix}\)
\(\Rightarrow\)\(\frac{1}{2}\)[3(3-y)-1(9-x)+1(9y-3x)]
\(\Rightarrow\) 9-3y-9+x+9y-3x=0
\(\Rightarrow\) x-3y=0
Hence, the equation of the line joining the given points is x − 3y = 0
If \(\begin{vmatrix} 2x & 3 \\ x & -8 \\ \end{vmatrix} = 0\), then the value of \(x\) is:
Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to:
If $ y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\27 & 28 & 27 \\1 & 1 & 1 \end{vmatrix} $, $ x \in \mathbb{R} $, then $ \frac{d^2y}{dx^2} + y $ is equal to
