Writing Equations Of Lines Parallel And Perpendicular To A Given Line Through A Point

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:

$\begin{aligned} y &= m_1 x + b_1\\ y &= m_2 x + b_2. \end{aligned}$

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

$\begin{aligned} y &= \frac{1}{2} x + 0\\ y &= \frac{1}{2} x + 3. \end{aligned}$

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 $(3, 5)?$

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

$\begin{aligned} y &= m_1 x + b_1\\ y &= m_2 x + b_2, \end{aligned}$

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

$\begin{aligned} y &= \frac{1}{2} x + 3\\ y &= -2x -2, \end{aligned}$

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 $y=\frac{1}{5}x-2?$

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 $\begin{array}{c}&(k+1)x-3y+2=0, &(k-2)x+4y-1=0\end{array}$ are perpendicular to each other?

For the two lines to be perpendicular, it must be true that $(k+1)\times(k-2)+(-3)\times 4=0.$ Hence, $\begin{aligned} (k+1)\times(k-2)+(-3)\times 4&=0\\ k^2-k-2-12&=0\\ k^2-k-14&=0. \end{aligned}$ 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 $a+b?$

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 $a+b=-3-\frac{1}{2}=-\frac{7}{2}. \ _\square$