4.1 Linear functions  (Page 12/27)

 Page 12 / 27
$-4\left(\frac{1}{4}\right)=-1$

Parallel and perpendicular lines

Two lines are parallel lines    if they do not intersect. The slopes of the lines are the same.

If and only if $\text{\hspace{0.17em}}{b}_{1}={b}_{2}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{m}_{1}={m}_{2},\text{\hspace{0.17em}}$ we say the lines coincide. Coincident lines are the same line.

Two lines are perpendicular lines    if they intersect to form a right angle.

${m}_{1}{m}_{2}=-1,\text{so}\text{\hspace{0.17em}}{m}_{2}=-\frac{1}{{m}_{1}}$

Identifying parallel and perpendicular lines

Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.

$\begin{array}{cccccc}\hfill f\left(x\right)& =& 2x+3\hfill & \hfill \phantom{\rule{2em}{0ex}}h\left(x\right)& =& -2x+2\hfill \\ \hfill g\left(x\right)& =& \frac{1}{2}x-4\hfill & \hfill \phantom{\rule{2em}{0ex}}j\left(x\right)& =& 2x-6\hfill \end{array}$

Parallel lines have the same slope. Because the functions $\text{\hspace{0.17em}}f\left(x\right)=2x+3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}j\left(x\right)=2x-6\text{\hspace{0.17em}}$ each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and $\text{\hspace{0.17em}}\frac{1}{2}\text{\hspace{0.17em}}$ are negative reciprocals, the functions $\text{\hspace{0.17em}}g\left(x\right)=\frac{1}{2}x-4\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h\left(x\right)=-2x+2\text{\hspace{0.17em}}$ represent perpendicular lines.

Writing the equation of a line parallel or perpendicular to a given line

If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.

Writing equations of parallel lines

Suppose for example, we are given the equation shown.

$f\left(x\right)=3x+1$

We know that the slope of the line formed by the function is 3. We also know that the y- intercept is $\text{\hspace{0.17em}}\left(0,1\right).\text{\hspace{0.17em}}$ Any other line with a slope of 3 will be parallel to $\text{\hspace{0.17em}}f\left(x\right).\text{\hspace{0.17em}}$ So the lines formed by all of the following functions will be parallel to $\text{\hspace{0.17em}}f\left(x\right).$

$\begin{array}{ccc}\hfill g\left(x\right)& =& 3x+6\hfill \\ \hfill h\left(x\right)& =& 3x+1\hfill \\ \hfill p\left(x\right)& =& 3x+\frac{2}{3}\hfill \end{array}$

Suppose then we want to write the equation of a line that is parallel to $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and passes through the point $\text{\hspace{0.17em}}\left(1,\text{7}\right).\text{\hspace{0.17em}}$ This type of problem is often described as a point-slope problem because we have a point and a slope. In our example, we know that the slope is 3. We need to determine which value of $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.

$\begin{array}{ccc}\hfill y-{y}_{1}& =& m\left(x-{x}_{1}\right)\hfill \\ \hfill y-7& =& 3\left(x-1\right)\hfill \\ \hfill y-7& =& 3x-3\hfill \\ \hfill y& =& 3x+4\hfill \end{array}$

So $\text{\hspace{0.17em}}g\left(x\right)=3x+4\text{\hspace{0.17em}}$ is parallel to $\text{\hspace{0.17em}}f\left(x\right)=3x+1\text{\hspace{0.17em}}$ and passes through the point $\text{\hspace{0.17em}}\left(1,\text{7}\right).$

Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.

1. Find the slope of the function.
2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.
3. Simplify.

Finding a line parallel to a given line

Find a line parallel to the graph of $\text{\hspace{0.17em}}f\left(x\right)=3x+6\text{\hspace{0.17em}}$ that passes through the point $\text{\hspace{0.17em}}\left(3,\text{0}\right).$

The slope of the given line is 3. If we choose the slope-intercept form, we can substitute $\text{\hspace{0.17em}}m=3,x=3,$ and $\text{\hspace{0.17em}}f\left(x\right)=0\text{\hspace{0.17em}}$ into the slope-intercept form to find the y- intercept.

$\begin{array}{ccc}\hfill g\left(x\right)& =& 3x+b\hfill \\ \hfill 0& =& 3\left(3\right)+b\hfill \\ \hfill b& =& –9\hfill \end{array}$

The line parallel to $\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$ that passes through $\text{\hspace{0.17em}}\left(3,\text{0}\right)\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}g\left(x\right)=3x-9.$

Writing equations of perpendicular lines

We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the function shown.

what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
Ajith
exponential series
Naveen
yeah
Morosi
prime number?
Morosi
what is subgroup
Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1