4.1 Linear functions  (Page 12/27)

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$-4\left(\frac{1}{4}\right)=-1$

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.

Cos45/sec30+cosec30=
Cos 45 = 1/ √ 2 sec 30 = 2/√3 cosec 30 = 2. =1/√2 / 2/√3+2 =1/√2/2+2√3/√3 =1/√2*√3/2+2√3 =√3/√2(2+2√3) =√3/2√2+2√6 --------- (1) =√3 (2√6-2√2)/((2√6)+2√2))(2√6-2√2) =2√3(√6-√2)/(2√6)²-(2√2)² =2√3(√6-√2)/24-8 =2√3(√6-√2)/16 =√18-√16/8 =3√2-√6/8 ----------(2)
exercise 1.2 solution b....isnt it lacking
I dnt get dis work well
what is one-to-one function
what is the procedure in solving quadratic equetion at least 6?
Almighty formula or by factorization...or by graphical analysis
Damian
I need to learn this trigonometry from A level.. can anyone help here?
yes am hia
Miiro
tanh2x =2tanhx/1+tanh^2x
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)=cotb ... pls some one should help me with this..thanks in anticipation
f(x)=x/x+2 given g(x)=1+2x/1-x show that gf(x)=1+2x/3
proof
AUSTINE
sebd me some questions about anything ill solve for yall
cos(a+b)+cos(a-b)/sin(a+b)-sin(a-b)= cotb
favour
how to solve x²=2x+8 factorization?
x=2x+8 x-2x=2x+8-2x x-2x=8 -x=8 -x/-1=8/-1 x=-8 prove: if x=-8 -8=2(-8)+8 -8=-16+8 -8=-8 (PROVEN)
Manifoldee
x=2x+8
Manifoldee
×=2x-8 minus both sides by 2x
Manifoldee
so, x-2x=2x+8-2x
Manifoldee
then cancel out 2x and -2x, cuz 2x-2x is obviously zero
Manifoldee
so it would be like this: x-2x=8
Manifoldee
then we all know that beside the variable is a number (1): (1)x-2x=8
Manifoldee
so we will going to minus that 1-2=-1
Manifoldee
so it would be -x=8
Manifoldee
so next step is to cancel out negative number beside x so we get positive x
Manifoldee
so by doing it you need to divide both side by -1 so it would be like this: (-1x/-1)=(8/-1)
Manifoldee
so -1/-1=1
Manifoldee
so x=-8
Manifoldee
Manifoldee
so we should prove it
Manifoldee
x=2x+8 x-2x=8 -x=8 x=-8 by mantu from India
mantu
lol i just saw its x²
Manifoldee
x²=2x-8 x²-2x=8 -x²=8 x²=-8 square root(x²)=square root(-8) x=sq. root(-8)
Manifoldee
I mean x²=2x+8 by factorization method
Kristof
I think x=-2 or x=4
Kristof
x= 2x+8 ×=8-2x - 2x + x = 8 - x = 8 both sides divided - 1 -×/-1 = 8/-1 × = - 8 //// from somalia
Mohamed
i am in
Cliff
hii
Amit
how are you
Dorbor
well
Biswajit
can u tell me concepts
Gaurav
Find the possible value of 8.5 using moivre's theorem
which of these functions is not uniformly cintinuous on (0, 1)? sinx
helo
Akash
hlo
Akash
Hello
Hudheifa
which of these functions is not uniformly continuous on 0,1