Area Bounded by the Lines: x0, y0, xy1, and 2x3y6
Introduction
The calculation of the area bounded by a set of lines provides a fundamental problem in coordinate geometry. This article aims to find the area of the region bounded by the lines x0, y0, xy1, and 2x-3y6. We will break down the process step-by-step, focusing on understanding the geometry and application of basic formulas.
Intercepts and Intersection Points
The first two lines, x0 and y0, represent the y-axis and x-axis, respectively. For the third line, xy1, we can rewrite it in two useful forms for analysis:
Equation 1: x 1/y (implying that the x-intercept is 1 when y1, and similarly for y-intercept). Equation 2: y 1/x (implying that the y-intercept is 1 when x1, and similarly for x-intercept).For the fourth line, 2x - 3y 6, we can also rewrite it in a useful form for understanding its intercepts:
Equation 3: y 1 - (2/3)x (implying that the y-intercept is 2 when x 0, and the x-intercept is 3 when y 0).These equations will help us visualize and compute the area of the region formed by their intersections.
Graphing and Analyzing the Line Equations
We start by graphing the y-axis (x0) and the x-axis (y0). These lines intersect at the origin (0,0).
The third line, 2x - 3y 6, intersects the x-axis at (3,0) and the y-axis at (0,2). From this, we can generate the line equation in the slope-intercept form:
slope (m) 2/3 y-intercept (b) 2 Line equation: y 2/3x 2The fourth line, xy 1, can be rewritten in slope-intercept form as:
Line equation: y 1/x In the first quadrant, y 1/x will form a hyperbolic curve that intersects the x-axis and y-axis at (1,1).By graphing these lines, we can identify the vertices of the region we are interested in.
Area Calculation
Step 1: Finding the Area of the Triangle Formed by the Fourth Line and the Axes
The fourth line, 2x - 3y 6, forms a right triangle with the x and y axes. The base of this triangle is along the x-axis from (0,0) to (3,0) and the height is along the y-axis from (0,0) to (0,2).
Area calculation:
Area 1/2 * base * height 1/2 * 3 * 2 3 square units.
Step 2: Finding the Area of the Triangle Formed by the Third Line and the Axes
The third line, xy 1 (or xy - 1 0 in line form), intersects the x-axis at (1,0) and the y-axis at (0,1). This forms another right triangle with the axes.
Area calculation:
Area 1/2 * base * height 1/2 * 1 * 1 0.5 square units.
Next, we need to subtract the area of the smaller triangle from the area of the larger triangle to get the area of the remaining region.
Final Area Calculation
The required area is the difference in areas of the triangles formed by the lines 2x - 3y 6 and xy 1 with the axes:
Area 3 - 0.5 2.5 square units.
Conclusion
In conclusion, the area bounded by the lines x0, y0, xy1, and 2x-3y6 is 2.5 square units. This process demonstrates the application of basic geometric principles and algebraic manipulation in solving complex area problems.
Note: This solution uses the properties of lines and basic geometric area formulas to find the region of intersection and compute the area accurately.