Question

Find an equation of the line satisfying the conditions. Write your answer in the slope-intercept form. Passes through \((3,-5)\) and is parallel to the line with equation \(2 x+3 y=12\)

Short answer

The equation of the line that passes through the point \((3, -5)\) and is parallel to the line with the equation \(2x + 3y = 12\) is: \(y = -\frac{2}{3}x - 3\)

Step-by-step solution

Find the slope of the given line

To obtain the slope, we need to rewrite the equation of the line \(2x + 3y = 12\) in the slope-intercept form (\(y = mx + b\)), where \(m\) is the slope and \(b\) is the y-intercept. Subtract \(2x\) from both sides: \(3y = -2x + 12\) Divide both sides by \(3\): \(y = -\frac{2}{3}x + 4\) Now, the given line is in the slope-intercept form. The slope of this line is \(-\frac{2}{3}\).

Write the equation of the parallel line

Since parallel lines have the same slope, the slope of the line we seek is also \(-\frac{2}{3}\). Now we can set up the equation of the new line using the slope-intercept form: \(y = -\frac{2}{3}x + b\) We need to determine the y-intercept (\(b\)) by substituting the coordinates of the given point \((3, -5)\) into the equation: \(-5 = -\frac{2}{3}(3) + b\)

Solve for the y-intercept \(b\)

Now, we can solve for \(b\): \(-5 = -\frac{2}{3}(3) + b\) \(-5 = -2 + b\) Add \(2\) to both sides: \(-3 = b\)

Write the final equation

Now that we have the slope and the y-intercept, we can write the equation of the line in slope-intercept form: \(y = -\frac{2}{3}x - 3\) The equation of the line that passes through the point \((3, -5)\) and is parallel to the line with the equation \(2x + 3y = 12\) is: \(y = -\frac{2}{3}x - 3\)

Key concepts

Slope-Intercept Form

Understanding the slope-intercept form is critical when dealing with linear equations. It is written as \(y = mx + b\), where \(m\) represents the slope of the line, and \(b\) is the y-intercept. The slope indicates the steepness and direction of the line—'rise over run'. The y-intercept is where the line crosses the y-axis, indicating the point at which the line would be if \(x = 0\).

When you're given a linear equation not in this form, like \(2x + 3y = 12\), rewriting it to slope-intercept form reveals the slope and y-intercept directly. This step simplifies graphing and comparative analysis of lines, such as finding parallel or perpendicular lines.

Parallel Lines

Lines are parallel if they have the same slope, meaning they run alongside each other without ever crossing. If given the equation of one line, like \(2x + 3y = 12\), finding a line parallel to it simply requires using the same slope.

Keep in mind, although parallel lines share the same slope, their y-intercepts can be different, which translates into their differing positions on the graph. In our example, the original and new line are parallel because they both have a slope of \(-\frac{2}{3}\), yet their y-intercepts differ, placing them at unique locations on the coordinate plane.

Solving for y-Intercept

Solving for the y-intercept, designated as \(b\) in slope-intercept form, can be achieved by plugging in the coordinates of a known point on the line into the equation \(y = mx + b\).

Let's illustrate this with the point \((3, -5)\). Substituting into \(y = -\frac{2}{3}x + b\) yields \(-5 = -\frac{2}{3}(3) + b\), and solving for \(b\) gives us the y-intercept. This step is a cornerstone in defining the exact position of a line on a graph, as the y-intercept specifies the starting point of a line on the y-axis.

Graphing Linear Equations

Graphing linear equations becomes intuitive when using the slope-intercept form. Start by plotting the y-intercept \(b\) on the y-axis. Next, use the slope \(m\) to find another point on the line by 'rising' and 'running' from the y-intercept. A slope of \(-\frac{2}{3}\) means you drop down 2 units (since it's negative) and move 3 units to the right on the graph.

Join the y-intercept and this new point with a straight line, and you've graphed your equation. You can add more points using the slope for accuracy or to extend the line across the graph.