When finding the area of a triangle with vertices on the coordinate axes, use the formula for the area of a right triangle: \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). The base and height are the distances along the \( x \)-axis and \( y \)-axis, respectively, which can be found by determining the intercepts of the line with the axes. Simplify the expression carefully to find the correct area.
The equation of the line is:
\[ \frac{x}{7\sqrt{3a}} + \frac{y}{b} = 4. \]
Rewriting:
\[ b \cdot x + 7\sqrt{3a} \cdot y = 28\sqrt{3ab}. \]
To find the intercepts:
1. When \( x = 0 \), \( y = 4b \).
2. When \( y = 0 \), \( x = 28\sqrt{3a} \).
The lines \( x = 0 \) and \( y = 0 \), together with \( \frac{x}{7\sqrt{3a}} + \frac{y}{b} = 4 \), form a triangle with vertices:
\[ (0, 0), \quad (28\sqrt{3a}, 0), \quad (0, 4b). \]
The area of the triangle is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}, \]
where the base is \( 28\sqrt{3a} \) and the height is \( 4b \):
\[ \text{Area} = \frac{1}{2} \times (28\sqrt{3a}) \times (4b). \]
Simplify:
\[ \text{Area} = \frac{1}{2} \times 112\sqrt{3ab} = 56\sqrt{3ab}. \]
Thus, the area of the region is:
\[ 56\sqrt{3ab}. \]
The equation of the line is:
\[ \frac{x}{7\sqrt{3a}} + \frac{y}{b} = 4. \]
Step 1: Rewrite the equation:
Multiply the entire equation by \( b \cdot 7\sqrt{3a} \) to eliminate the fractions: \[ b \cdot x + 7\sqrt{3a} \cdot y = 28\sqrt{3ab}. \]Step 2: Find the intercepts:
To find the intercepts, we substitute \( x = 0 \) for the \( y \)-intercept and \( y = 0 \) for the \( x \)-intercept: - When \( x = 0 \), we have \( y = 4b \). - When \( y = 0 \), we have \( x = 28\sqrt{3a} \).Step 3: Form the triangle:
The lines \( x = 0 \), \( y = 0 \), and the equation \( \frac{x}{7\sqrt{3a}} + \frac{y}{b} = 4 \) form a triangle with the vertices at: \[ (0, 0), \quad (28\sqrt{3a}, 0), \quad (0, 4b). \]Step 4: Calculate the area of the triangle:
The area of a triangle is given by: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}. \] Here, the base is \( 28\sqrt{3a} \) and the height is \( 4b \). Therefore, the area is: \[ \text{Area} = \frac{1}{2} \times (28\sqrt{3a}) \times (4b). \] Simplify: \[ \text{Area} = \frac{1}{2} \times 112\sqrt{3ab} = 56\sqrt{3ab}. \]Conclusion: Thus, the area of the region is:
\[ 56\sqrt{3ab}. \]Fit a straight-line trend by the method of least squares for the following data:
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Year} & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 \\ \hline \textbf{Profit (₹ 000)} & 114 & 130 & 126 & 144 & 138 & 156 & 164 \\ \hline \end{array} \]List-I | List-II | ||
A | Megaliths | (I) | Decipherment of Brahmi and Kharoshti |
B | James Princep | (II) | Emerged in first millennium BCE |
C | Piyadassi | (III) | Means pleasant to behold |
D | Epigraphy | (IV) | Study of inscriptions |