Equations of Parallel and Perpendicular Lines

As we have seen in the wiki Slopes and Intercepts of a Line, every line in the \(xy\)-plane has a slope, which can be thought of as the rate of change of \(y\) with respect to \(x\). Given two lines, comparing the slopes of the two lines gives us important information about how the lines meet in the \(xy\)-plane (if at all).

Parallel lines are lines that do not meet at any point in the \(xy\)-plane. Another way to characterize parallel lines are distinct lines with the same slope. Suppose we are given two non-vertical lines in slope-intercept form:

\[ \begin{align} y &= m_1 x + b_1\\ y &= m_2 x + b_2. \end{align} \]

Then the two lines are parallel if \(m_1 = m_2\) and \(b_1 \ne b_2\).

Intuitively, if two distinct lines have the same rate of change, then the lines always point in the same direction and thus will never meet. In the above image, the slope-intercept form for the two lines are

\[ \begin{align} y &= \frac{1}{2} x + 0\\ y &= \frac{1}{2} x + 3. \end{align} \]

Since the two lines have the same slope and different \(y\)-intercepts, the two lines are parallel.

What is the equation of the line that is parallel to the line \(2x-3y-8=0\) and passes through the point \((3, 5)?\)

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Let \(y=ax+b\) be the equation of the line of interest. Then since this line is parallel to the line \(2x-3y-8=0\) or \(y=\frac{2x}{3}-\frac{8}{3},\) the slope of which is \(\frac{2}{3},\) so it must be true that \(a=\frac{2}{3}.\) So, the equation now becomes \(y=\frac{2}{3}x+b.\) Substituting in the coordinates \((3, 5),\) we have \[5=\frac{2}{3}\times 3+b \implies b=3.\] Therefore, the equation of the line of interest is \[y=\frac{2}{3}x+3. \ _\square\]

A pair of lines is perpendicular if the lines meet at \(90^\circ\) angle. Given two non-vertical lines in slope-intercept form

\[ \begin{align} y &= m_1 x + b_1\\ y &= m_2 x + b_2, \end{align} \]

the two lines are perpendicular if \(m_1 = - \frac{1}{m_2}\), that is, if the slopes are negative reciprocals of each other:

In the above image, the slope-intercept form of the two lines are

\[ \begin{align} y &= \frac{1}{2} x + 3\\ y &= -2x -2, \end{align} \]

and since the two slopes are negative reciprocals of each other, the lines are perpendicular.

What is the equation of the line that passes through the point \((-7, 3)\) and is perpendicular to the line \(y=\frac{1}{5}x-2?\)

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Let \(y=ax+b\) be the equation of the line of interest. Then since this line is perpendicular to the line \(y=\frac{1}{5}x-2\) the slope of which is \(\frac{1}{5},\) it must be true that \(a=-5.\) So, the equation now becomes \(y=-5x+b.\) Substituting in the coordinates \((-7, 3),\) we have \[3=-5\times (-7)+b \implies b=-32.\] Therefore, the equation of the line of interest is \[y=-5x-32. \ _\square\]

What is the sum of all the constants \(k\) such that the two lines \[\begin{array} &(k+1)x-3y+2=0, &(k-2)x+4y-1=0\end{array}\] are perpendicular to each other?

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For the two lines to be perpendicular, it must be true that \[(k+1)\times(k-2)+(-3)\times 4=0.\] Hence, \[\begin{align} (k+1)\times(k-2)+(-3)\times 4&=0\\ k^2-k-2-12&=0\\ k^2-k-14&=0. \end{align}\] Therefore, by Vieta's formula the sum of all the possible values of \(k\) is \(1.\) \(_\square\)

In some problems, we may be given properties of the slopes and intercepts of two lines and wish to calculate the values for the slopes and intercepts.

Consider two lines \(y=-2x+3\) and \(y=(K+1)x+4.\) When \(K=a,\) the two lines are parallel. When \(K=b,\) the two lines are perpendicular. What is \(a+b?\)

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Observe that the slope of the line \(y=-2x+3\) is \(-2\) and the slope of the line \(y=(K+1)x+4\) is \(K+1.\)

Then, since the two lines are parallel when \(K=a,\) it follows that \[a+1=-2 \implies a=-3.\] Similarly, since the two lines are perpendicular when \(K=b,\) it follows that \[b+1=\frac{1}{2} \implies b=-\frac{1}{2}.\] Therefore, our answer is \[a+b=-3-\frac{1}{2}=-\frac{7}{2}. \ _\square\]