b. Find the image of a point $$ A(3,4) $$ after reflection in $$ x $$-axis. 5.a. What is the formula of mean deviation from median in discrete series. Group- $$ B[8 imes 2=16] $$ b. Define transpose matrix of $$ A $$. 6.a. Define Rectangular matrix with example. b. If $$ A=\left[\begin{array}{ll}1 & 3 \ 2 & 4\end{array} ight] $$ and $$ B=\left[\begin{array}{ll}3 & 2 \ 6 & 4\end{array} ight] $$ then find $$ A B $$. 7.a. Find the distance between the points $$ A(-2,1) $$ and $$ B(4,3) $$. b. For what value of $$ K $$ will be the point $$ (2,1) $$ lie on the curve. $$ k x^{2}-2 y^{2}-2 x+3 y-3=0 $$ ? 8.a. Find the mid - point of the line joining the points $$ (-3,6) $$ and $$ (1,-2) $$ b. Prove that: $$ \left(1-\cos ^{2} heta ight) $$. $$ \left(1+\cot ^{2} heta ight)=1 $$ 9.a. If $$ A(1,3) $$ and $$ B(6,3) $$ are the two end points of $$ \overrightarrow{A B} $$. Find $$ \overrightarrow{A B} $$ and hence find $$ |\overrightarrow{A B}| $$.

Question

b. Find the image of a point $$ A(3,4) $$ after reflection in $$ x $$-axis. 5.a. What is the formula of mean deviation from median in discrete series. Group- $$ B[8 	imes 2=16] $$ b. Define transpose matrix of $$ A $$. 6.a. Define Rectangular matrix with example. b. If $$ A=\left[\begin{array}{ll}1 & 3 \ 2 & 4\end{array}ight] $$ and $$ B=\left[\begin{array}{ll}3 & 2 \ 6 & 4\end{array}ight] $$ then find $$ A B $$. 7.a. Find the distance between the points $$ A(-2,1) $$ and $$ B(4,3) $$. b. For what value of $$ K $$ will be the point $$ (2,1) $$ lie on the curve. $$ k x^{2}-2 y^{2}-2 x+3 y-3=0 $$ ? 8.a. Find the mid - point of the line joining the points $$ (-3,6) $$ and $$ (1,-2) $$ b. Prove that: $$ \left(1-\cos ^{2} 	hetaight) $$. $$ \left(1+\cot ^{2} 	hetaight)=1 $$ 9.a. If $$ A(1,3) $$ and $$ B(6,3) $$ are the two end points of $$ \overrightarrow{A B} $$. Find $$ \overrightarrow{A B} $$ and hence find $$ |\overrightarrow{A B}| $$.

UpStudy AI · Accepted Answer

**b. Image of Point $$ A(3,4) $$ after Reflection in the $$ x $$-axis**  
When a point is reflected in the $$ x $$-axis, its $$ x $$-coordinate remains the same and its $$ y $$-coordinate changes sign.  
Thus, the image of $$ A(3,4) $$ is  
$$
(3,-4)
$$

**5.a. Formula for Mean Deviation from the Median (Discrete Series)**  
For a discrete frequency distribution with values $$ x_1,x_2,\ldots,x_n $$ and corresponding frequencies $$ f_1,f_2,\ldots,f_n $$, and with median denoted by $$ \tilde{x} $$, the mean deviation from the median is given by  
$$
\text{Mean Deviation} = \frac{\sum_{i=1}^{n} f_i \left| x_i - \tilde{x} \right|}{\sum_{i=1}^{n} f_i}
$$

**5.b. Definition of the Transpose Matrix of $$ A $$**  
If  
$$
A = \begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n}\[6mm]
a_{21} & a_{22} & \cdots & a_{2n}\[6mm]
\vdots & \vdots & \ddots & \vdots\[6mm]
a_{m1} & a_{m2} & \cdots & a_{mn}
\end{bmatrix},
$$  
then the transpose of $$ A $$, denoted $$ A^T $$, is the matrix obtained by interchanging its rows and columns:  
$$
A^T = \begin{bmatrix}
a_{11} & a_{21} & \cdots & a_{m1}\[6mm]
a_{12} & a_{22} & \cdots & a_{m2}\[6mm]
\vdots & \vdots & \ddots & \vdots\[6mm]
a_{1n} & a_{2n} & \cdots & a_{mn}
\end{bmatrix}.
$$

**6.a. Definition of a Rectangular Matrix (with Example)**  
A rectangular matrix is a matrix in which the number of rows is not equal to the number of columns. For example,  
$$
\begin{bmatrix}
1 & 2 & 3\[4mm]
4 & 5 & 6
\end{bmatrix}
$$  
is a rectangular matrix with $$ 2 $$ rows and $$ 3 $$ columns.

**6.b. Finding $$ AB $$ for Given Matrices**  
Given  
$$
A = \begin{bmatrix} 1 & 3 \[4mm] 2 & 4 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 3 & 2 \[4mm] 6 & 4 \end{bmatrix},
$$  
the product $$ AB $$ is calculated by taking the dot product of rows of $$ A $$ with columns of $$ B $$:

- Entry $$(1,1)$$: $$1 \times 3 + 3 \times 6 = 3 + 18 = 21$$.  
- Entry $$(1,2)$$: $$1 \times 2 + 3 \times 4 = 2 + 12 = 14$$.  
- Entry $$(2,1)$$: $$2 \times 3 + 4 \times 6 = 6 + 24 = 30$$.  
- Entry $$(2,2)$$: $$2 \times 2 + 4 \times 4 = 4 + 16 = 20$$.

Thus,  
$$
AB = \begin{bmatrix} 21 & 14 \[4mm] 30 & 20 \end{bmatrix}.
$$

**7.a. Distance Between Points $$ A(-2,1) $$ and $$ B(4,3) $$**  
Using the distance formula,  
$$
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2},
$$  
we substitute $$ A(-2,1) $$ and $$ B(4,3) $$:  
$$
d = \sqrt{(4 - (-2))^2 + (3 - 1)^2} 
= \sqrt{(6)^2 + (2)^2} 
= \sqrt{36+4} 
= \sqrt{40}
= 2\sqrt{10}.
$$

**7.b. Finding $$ k $$ for the Point $$ (2,1) $$ to Lie on the Curve**  
The curve is given by  
$$
kx^{2}-2y^{2}-2x+3y-3=0.
$$  
Substitute $$ x=2 $$ and $$ y=1 $$:  
$$
k(2)^2 - 2(1)^2 - 2(2) + 3(1) - 3 = 0.
$$  
Simplify:  
$$
4k - 2 - 4 + 3 - 3 = 0,
$$  
$$
4k - 6 = 0.
$$  
Solving for $$ k $$:  
$$
4k = 6 \quad \Longrightarrow \quad k = \frac{6}{4} = \frac{3}{2}.
$$

**8.a. Midpoint of the Line Joining $$ (-3,6) $$ and $$ (1,-2) $$**  
The midpoint $$ M $$ is given by  
$$
M = \left( \frac{x_1+x_2}{2},\, \frac{y_1+y_2}{2} \right).
$$  
For $$ (-3,6) $$ and $$ (1,-2) $$:  
$$
M = \left( \frac{-3+1}{2},\, \frac{6+(-2)}{2} \right) 
= \left( \frac{-2}{2},\, \frac{4}{2} \right) 
= (-1,2).
$$

**8.b. Proving $$ \left( 1 -\cos^{2} \theta \right) \left( 1 + \cot^{2} \theta \right) = 1 $$**  
We start with the trigonometric identities:  
- $$ 1 - \cos^{2} \theta = \sin^{2} \theta $$ (using $$ \sin^2\theta + \cos^2\theta = 1 $$).  
- $$ 1 + \cot^{2} \theta = \csc^{2} \theta $$ (a standard Pythagorean identity).

Multiplying these we get:  
$$
\left( 1 - \cos^{2} \theta \right)\left( 1 + \cot^{2} \theta \right)
= \sin^{2} \theta \cdot \csc^{2} \theta.
$$  
Since $$ \csc \theta = \frac{1}{\sin \theta} $$, we have  
$$
\sin^{2} \theta \cdot \csc^{2} \theta = \sin^{2} \theta \cdot \left(\frac{1}{\sin \theta}\right)^2 
= \sin^{2} \theta \cdot \frac{1}{\sin^{2} \theta} 
= 1.
$$

**9.a. Finding $$ \overrightarrow{AB} $$ and Its Magnitude**  
Given the endpoints $$ A(1,3) $$ and $$ B(6,3) $$, the vector $$ \overrightarrow{AB} $$ is calculated by subtracting the coordinates of $$ A $$ from $$ B $$:  
$$
\overrightarrow{AB} = (6-1,\, 3-3) = (5,0).
$$  
Its magnitude is given by:  
$$
|\overrightarrow{AB}| = \sqrt{5^2+0^2} = \sqrt{25} = 5.
$$
**b. Image of Point $$ A(3,4) $$ after Reflection in the $$ x $$-axis:** $$ (3, -4) $$ **5.a. Mean Deviation from the Median (Discrete Series):** $$ \text{Mean Deviation} = \frac{\sum_{i=1}^{n} f_i |x_i - \tilde{x}|}{\sum_{i=1}^{n} f_i} $$ **5.b. Transpose of Matrix $$ A $$:** Interchange the rows and columns of $$ A $$. **6.a. Rectangular Matrix:** A matrix with unequal numbers of rows and columns, e.g., $$ \begin{bmatrix} 1 & 2 & 3\ 4 & 5 & 6 \end{bmatrix} $$ **6.b. Product $$ AB $$:** $$ AB = \begin{bmatrix} 21 & 14 \ 30 & 20 \end{bmatrix} $$ **7.a. Distance Between $$ A(-2,1) $$ and $$ B(4,3) $$:** $$ 2\sqrt{10} $$ **7.b. Value of $$ k $$:** $$ k = \frac{3}{2} $$ **8.a. Midpoint of $$ (-3,6) $$ and $$ (1,-2) $$:** $$ (-1, 2) $$ **8.b. Proof:** $$ (1 - \cos^2 \theta)(1 + \cot^2 \theta) = 1 $$ **9.a. Vector $$ \overrightarrow{AB} $$ and Its Magnitude:** $$ \overrightarrow{AB} = (5, 0), \quad |\overrightarrow{AB}| = 5 $$

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