Problem 5 - College Algebra

Question

Find the area of a parallelogram bounded by the \(y\) -axis, the line \(x=3,\) the line \(f(x)=1+2 x,\) and the line parallel to \(f(x)\) passing through \((2,7)\)

Step-by-Step Solution

Answer

The area of the parallelogram is 6 square units.

Step 1: Understand the Problem

We are asked to find the area of a parallelogram bounded by four lines: the y-axis (x = 0), the line x = 3, the line f(x) = 1 + 2x, and another line parallel to f(x) that passes through the point (2,7). A parallelogram formed by these lines will have a base and a height that we need to determine.

Step 2: Find the Equation of the Parallel Line

Since we need a line parallel to f(x) = 1 + 2x that passes through the point (2, 7), we start with the known slope of 2 (as parallel lines share the same slope). Using the point-slope form, the equation is y - 7 = 2(x - 2). Simplifying, we get y = 2x + 3.

Step 3: Identify the Points of Intersection with the y-axis

The line f(x) = 1 + 2x intersects the y-axis when x = 0, giving the point (0, 1). The parallel line y = 2x + 3 intersects the y-axis at x = 0, giving the point (0, 3). These two points (0,1) and (0,3) are endpoints of one side of the parallelogram.

Step 4: Identify the Points of Intersection with the Line x=3

For the line x = 3, f(x) = 1 + 2(3) = 7, so the intersection point is (3, 7). For the parallel line y = 2x + 3, substitute x = 3 to get y = 2(3) + 3 = 9. Thus, the intersection point is (3, 9). These points (3,7) and (3,9) form another side of the parallelogram.

Step 5: Calculate the Base and Height of the Parallelogram

The base is the vertical distance between the two intersections with the same x-coordinate, either at y=1 to y=3 or at y=7 to y=9, both of which are 2 units. The height is the horizontal distance between the lines intersecting with the y-axis and with the line x=3, which is 3 units (from x = 0 to x = 3).

Step 6: Calculate the Area

The area of a parallelogram is given by the formula: Area = Base × Height. In this case, the base is 2 units and the height is 3 units, so the area is 2 × 3 = 6 square units.

Key Concepts

Parallelogram Area

Area calculation is an essential concept in geometry, particularly for shapes like parallelograms. Remember that a parallelogram is a four-sided figure with opposite sides that are parallel and equal in length.
Finding the area involves knowing the base and the height. The base is any one of the sides, while the height is the perpendicular distance from the base to the opposite side.
Key steps to calculate the area include:

• Identify the base and height. Ensure they are perpendicular.
• Use the formula: Area = Base × Height.

In our exercise, we identified the base and height as 2 units and 3 units, respectively, resulting in an area of 6 square units. Understanding the area helps in applications in physics, engineering, and architecture, where space is a factor.

Lines and Slopes

Lines and their slopes are crucial in algebra, especially when working with equations. The slope measures the steepness of a line and is calculated as \( \frac{\text{rise}}{\text{run}} \), or \( \frac{\Delta y}{\Delta x} \).
Parallel lines share the same slope, making it easy to spot when lines are parallel. Perpendicular lines, meanwhile, have slopes that are negative reciprocals of each other.

• For a line equation like \( y = mx + b \), "m" represents the slope.
• A larger "m" indicates a steeper line.

In the given problem, both lines \( y = 1 + 2x \) and its parallel line \( y = 2x + 3 \) have a slope of 2, thus confirming they are parallel.

Equation of a Line

Equations of lines are typically expressed in the form \( y = mx + b \), where "m" is the slope and "b" is the y-intercept.

• To find an equation for a line parallel to another, use the same slope.
• Use the point-slope form: \( y - y_1 = m(x - x_1) \), to find the equation through any given point \((x_1, y_1)\).

In the exercise, we derived the equation of the line parallel to \( y = 1 + 2x \) that passes through the point (2, 7). Using the point-slope formula, the equation simplifies to \( y = 2x + 3 \). Understanding line equations is key to solving geometric problems.

Points of Intersection

Points of intersection are where two lines meet on a graph. Finding these points involves solving equations simultaneously. These intersections are critical for determining a parallelogram's vertices.
In the exercise, you needed to determine where the lines:

• intersect with the y-axis.
• intersect with a vertical line (e.g., \( x = 3 \)).

For a line \( y = mx + b \), plugging in \( x = 0 \) gives a y-coordinate, showing where it meets the y-axis. Similarly, plug in another \( x \)-value to find another intersection point.
In our problem, the intersections were (0, 1) and (0, 3) on the y-axis, and (3, 7) and (3, 9) on \( x = 3 \). Understanding points of intersection aids in visualizing and solving for areas and other geometric properties.