# Parallel And Perpendicular Worksheet Algebra 1 5.6 Homework Answer Key

## How to Solve Problems Involving Parallel and Perpendicular Lines: Algebra 1 5.6 Homework Answer Key

In this article, we will review some concepts and skills related to parallel and perpendicular lines in algebra. We will also provide some examples and solutions from the Algebra 1 5.6 homework worksheet.

## parallel and perpendicular worksheet algebra 1 5.6 homework answer key

## What are Parallel and Perpendicular Lines?

Parallel lines are lines in the same plane that never intersect. They have the same slope, which means they rise and run at the same rate. For example, the lines y = 2x + 5 and y = 2x - 3 are parallel because they both have a slope of 2.

Perpendicular lines are lines in the same plane that intersect at right angles (90 degrees). They have slopes that are negative reciprocals, which means they are opposite and flipped. For example, the lines y = -3x + 2 and y = (1/3)x - 4 are perpendicular because their slopes are -3 and (1/3), respectively. To find the slope of a perpendicular line, you can use this formula: m = -1/m, where m is the slope of the given line and m is the slope of the perpendicular line.

## How to Find Equations of Parallel and Perpendicular Lines?

To find the equation of a line that is parallel or perpendicular to a given line, you need to know two things: the slope of the line and a point on the line. You can use the point-slope form of a linear equation to write the equation of the line: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point.

If you want to find a parallel line, you just use the same slope as the given line. If you want to find a perpendicular line, you use the negative reciprocal of the slope of the given line. You can then simplify the equation by distributing and rearranging terms.

## Examples and Solutions from Algebra 1 5.6 Homework Worksheet

Here are some problems from the Algebra 1 5.6 homework worksheet on parallel and perpendicular lines, along with their solutions and explanations.

Find an equation of a line that passes through (4, -2) and is parallel to y = -5x + 3.

Solution: Since we want a parallel line, we use the same slope as the given line, which is -5. Then we plug in the slope and the point into the point-slope form: y - (-2) = -5(x - 4). Simplifying, we get y + 2 = -5x + 20, or y = -5x + 18.

Find an equation of a line that passes through (-3, 5) and is perpendicular to 3x - 7y = 21.

Solution: Since we want a perpendicular line, we need to find the negative reciprocal of the slope of the given line. To do that, we first solve for y and put the given equation in slope-intercept form: y = (3/7)x - 3. The slope of this line is (3/7), so the negative reciprocal is -(7/3). Then we plug in the slope and the point into the point-slope form: y - 5 = -(7/3)(x + 3). Simplifying, we get y - 5 = -(7/3)x - 7, or y = -(7/3)x - 2.

Determine if the lines given by x + y = 4 and x - y = -6 are parallel, perpendicular, or neither.

Solution: To compare the slopes of these lines, we need to solve for y and put them in slope-intercept form: y = -x + 4 and y = x + 6. The slopes of these lines are -1 and 1, respectively. Since these slopes are negative reciprocals, we conclude that these lines are perpendicular.

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