Question

In Exercises \(31-34,\) write an equation for the line through \(P\) that is (a) parallel to \(L,\) and (b) perpendicular to \(L .\) $$P(0,0), \quad L : y=-x+2$$

Short answer

The equation of the line through P(0,0) parallel to L is \(y = -x\). The equation of the line perpendicular to L is \(y = x\).

Step-by-step solution

Solving for the Line Parallel to L

We already know that parallel lines have the same slope. Thus, the line that is parallel to L also has a slope of -1. Remembering the equation for a line in slope-intercept form is \(y = mx + c\), where m is the slope and c is the y-intercept. We know the slope, m is -1 and it passes through the origin (0,0). Therefore, the equation of the line is \(y = -x + c\). Using point P(0,0) into \(y = -x + c\), we get \(0 = -0 + c\). So c = 0 and the equation of the line parallel to L that passes through P becomes: \(y = -x\).

Calculating the perpendicular line

The slope of a perpendicular line is the negative reciprocal of the original line's slope. Our line L has a slope of -1, so the negative reciprocal is 1. Using equation \(y = mx + c\) again, where m is now our new slope 1. As this line also passes through P(0,0), substituting into \(y = x + c\), we get \(0 = 0 + c\). So c = 0. So, the equation of the line perpendicular to L that passes through P is \(y = x\).

Key concepts

Slope-Intercept Form

The slope-intercept form of a line is a way to express the equation of a line. It is written as \(y = mx + c\). In this representation, \(m\) stands for the slope of the line, which measures how steep the line is, while \(c\) represents the y-intercept, which is the point where the line crosses the y-axis. This form is especially useful because it is straightforward to plot a line when we know these two characteristics.

For instance, if you are given a line with the equation \(y = 2x + 3\), you can immediately tell that the slope, or steepness, of the line is 2, and the line will cross the y-axis at \((0,3)\). Understanding how to work with slope-intercept form equations is essential in graphing lines and understanding their properties.

Parallel Line Equation

Parallel lines are lines in a plane that never meet; they have the same slope, but different y-intercepts. To determine the equation of a line that is parallel to another line, you first need to know the slope of the original line.

Identifying a Parallel Line

If you're given an equation in slope-intercept form, like \(y = -x + 2\), the slope is the coefficient of \(x\), which is -1 in this case. A parallel line would have the same slope, so its equation will start with \(y = -x + c\), where \(c\) can be any number representing the y-intercept. If the parallel line passes through a specific point, you use that point to solve for \(c\).

Perpendicular Line Equation

Perpendicular lines intersect at a right angle, creating an 'L' shape. The slopes of two perpendicular lines are negative reciprocals of each other. This means that if one line has a slope of \(a\), the other line will have a slope of \(-\frac{1}{a}\), provided that \(a\) is not equal to zero.

Deriving the Perpendicular Slope

For the line with equation \(y = -x + 2\), which has a slope of -1, the perpendicular line will have a slope of 1, the negative reciprocal of -1. So, if this line also passes through the origin as in our example problem, its equation would be \(y = x\), where the y-intercept is also 0.

Negative Reciprocal

The concept of the negative reciprocal is crucial when working with perpendicular lines. The negative reciprocal of a number is found by inverting the number and then changing its sign. If you have a fraction, you flip the numerator and denominator; if the original number is a whole number or an integer, you treat it as over 1 and then flip it.

For example, the negative reciprocal of 2 (which is the same as \(\frac{2}{1}\)) is \(-\frac{1}{2}\). Likewise, the negative reciprocal of -3 is \(\frac{1}{3}\). In the context of lines, if you know the slope of one line, the negative reciprocal will give you the slope of a perpendicular line.