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:
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
A school is organizing a debate competition with participants as speakers and judges. $ S = \{S_1, S_2, S_3, S_4\} $ where $ S = \{S_1, S_2, S_3, S_4\} $ represents the set of speakers. The judges are represented by the set: $ J = \{J_1, J_2, J_3\} $ where $ J = \{J_1, J_2, J_3\} $ represents the set of judges. Each speaker can be assigned only one judge. Let $ R $ be a relation from set $ S $ to $ J $ defined as: $ R = \{(x, y) : \text{speaker } x \text{ is judged by judge } y, x \in S, y \in J\} $.