The area of the region bounded by the lines $ \frac{x}{7\sqrt{3}a} + \frac{y}{b} = 4 $, $x=0$ and $y=0$ is :

Question

Prepp Team · Accepted Answer

Understanding the Problem: Area Bounded by a Line and Axes

The question asks us to find the area of the geometric region enclosed by a specific straight line and the two coordinate axes. The lines given are:

The line: $ \frac{x}{7\sqrt{3}a} + \frac{y}{b} = 4 $
The y-axis: $ x = 0 $
The x-axis: $ y = 0 $

The region bounded by a straight line and the x and y axes (where the line intersects both axes in the positive part, assuming $a$ and $b$ are positive constants) is typically a right-angled triangle. The vertices of this triangle are the origin (0,0), the point where the line crosses the x-axis (x-intercept), and the point where the line crosses the y-axis (y-intercept).

Finding the Intercepts of the Line

To find the area of the triangular region, we need to determine the lengths of the base and height. These correspond to the x and y intercepts of the given line.

Calculating the x-intercept

The x-intercept is the point where the line crosses the x-axis. On the x-axis, the y-coordinate is always 0. So, we set $ y = 0 $ in the equation of the line:

$ \frac{x}{7\sqrt{3}a} + \frac{0}{b} = 4 $

This simplifies to:

$ \frac{x}{7\sqrt{3}a} = 4 $

Solving for $ x $:

$ x = 4 \times 7\sqrt{3}a $

$ x = 28\sqrt{3}a $

So, the line intersects the x-axis at the point $ (28\sqrt{3}a, 0) $. The length of the base of our triangle along the x-axis is $ 28\sqrt{3}a $.

Calculating the y-intercept

The y-intercept is the point where the line crosses the y-axis. On the y-axis, the x-coordinate is always 0. So, we set $ x = 0 $ in the equation of the line:

$ \frac{0}{7\sqrt{3}a} + \frac{y}{b} = 4 $

This simplifies to:

$ \frac{y}{b} = 4 $

Solving for $ y $:

$ y = 4b $

So, the line intersects the y-axis at the point $ (0, 4b) $. The length of the height of our triangle along the y-axis is $ 4b $.

Calculating the Area of the Bounded Region

The region bounded by the line and the coordinate axes is a right-angled triangle with vertices at $ (0,0) $, $ (28\sqrt{3}a, 0) $, and $ (0, 4b) $. The lengths of the sides forming the right angle are the absolute values of the x and y intercepts. Assuming $ a > 0 $ and $ b > 0 $, the base is $ 28\sqrt{3}a $ and the height is $ 4b $.

The formula for the area of a triangle is:

$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $

Substituting the values of the base and height:

$ \text{Area} = \frac{1}{2} \times (28\sqrt{3}a) \times (4b) $

Now, we multiply the terms:

$ \text{Area} = \frac{1}{2} \times (28 \times 4) \times \sqrt{3} \times a \times b $

$ \text{Area} = \frac{1}{2} \times 112 \times \sqrt{3} ab $

$ \text{Area} = 56 \sqrt{3} ab $

Thus, the area of the region bounded by the given line and the coordinate axes is $ 56\sqrt{3} ab $.

Summary of Steps

Identify the bounding lines: the given line, $x=0$, and $y=0$.
Recognize that this region is a triangle (assuming intercepts are on positive axes).
Calculate the x-intercept by setting $y=0$ in the line equation. This gives the base length.
Calculate the y-intercept by setting $x=0$ in the line equation. This gives the height length.
Use the formula for the area of a triangle: $ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $.
Substitute the calculated base and height values and simplify.

Concept	Value/Formula
Line Equation	$ \frac{x}{7\sqrt{3}a} + \frac{y}{b} = 4 $
x-intercept (Base)	$ 28\sqrt{3}a $
y-intercept (Height)	$ 4b $
Area Formula	$ \frac{1}{2} \times \text{base} \times \text{height} $
Calculated Area	$ 56\sqrt{3} ab $

Revision Table: Key Concepts in Area Calculation

Term	Definition/Relevance
Line Equation	Mathematical representation of a straight line relating x and y coordinates. Can be in various forms (slope-intercept, standard, intercept form).
Coordinate Axes	The x-axis (where $y=0$) and the y-axis (where $x=0$). They form the basis of the Cartesian coordinate system.
x-intercept	The point where a line crosses the x-axis. Found by setting $y=0$ in the line equation. It's the distance from the origin along the x-axis.
y-intercept	The point where a line crosses the y-axis. Found by setting $x=0$ in the line equation. It's the distance from the origin along the y-axis.
Area of a Triangle	The measure of the surface enclosed by a triangle. For a right-angled triangle with legs along the axes, the area is half the product of the lengths of the legs (intercepts).
Region Bounded By Lines	The area enclosed within the boundaries defined by specific lines or curves. For a line and axes in the first quadrant, it's a triangle.

Additional Information: Lines and Area Calculation

Understanding the geometry of linear equations is fundamental. A linear equation like $ Ax + By = C $ or $ \frac{x}{a'} + \frac{y}{b'} = 1 $ represents a straight line. The intercept form, $ \frac{x}{a'} + \frac{y}{b'} = 1 $, directly shows the x-intercept $ a' $ and the y-intercept $ b' $. In our given equation, $ \frac{x}{7\sqrt{3}a} + \frac{y}{b} = 4 $, we can rewrite it in the intercept form by dividing by 4:

$ \frac{x}{4 \times 7\sqrt{3}a} + \frac{y}{4b} = 1 $

$ \frac{x}{28\sqrt{3}a} + \frac{y}{4b} = 1 $

From this form, the x-intercept is $ 28\sqrt{3}a $ and the y-intercept is $ 4b $. This confirms our earlier calculations.

The area bounded by the line $ \frac{x}{a'} + \frac{y}{b'} = 1 $ and the coordinate axes in the first quadrant is $ \frac{1}{2} |a' b'| $. This formula is a shortcut for calculating the area of the triangle formed by the origin, $(a', 0)$, and $(0, b')$. In our case, $ a' = 28\sqrt{3}a $ and $ b' = 4b $. Assuming $ a, b > 0 $, the area is:

$ \text{Area} = \frac{1}{2} (28\sqrt{3}a)(4b) = 56\sqrt{3} ab $

This method provides a quick way to verify the result obtained by explicitly finding the intercepts and using the triangle area formula.

Prepp Team · Answer

56a

Prepp Team · Answer

ab/2

Prepp Team · Answer

3ab