Question

Please Help.

1.Write the equation of the line which passes through (5, –2) and is parallel to x = 4.

2.Write the equation of the line which passes through (2, 1) and is perpendicular to x = –2.

3.Write the equation of the line which passes through (–4, 2) and is parallel to y = –x + 6 in slope-intercept form.

4.Write the equation of the line which passes through (2, –3) and is perpendicular to y = 4x + 7 in standard form.

Need to shw work

Answer 1

nothing to show for the first one \(x=4\) is a vertical line, so you are asked for the equation of another vertical line, this one through the point \((5,-2)\)
you write \(x=\text{the first coordinate}\)

Answer 2

For the first one is it x = 5

Answer 3

second one is similar, but since \(x=-2\) is a vertical line, you want the equation for a horizontal line
horizontal line looks like \(y=\text{some number}\) and since you want it to go through the point
\((2,1)\) you write \(y=\text{second coordinate}\)

Answer 4

yes, first one is \(x=5\)

Answer 5

you got the second one?

Answer 6

So y = 1

Answer 7

yes

Answer 8

third one requires some actual work
do you know what the slope of \(y=-x+6\) is ?

Answer 9

would the slope be a negative number

Answer 10

yes, in fact it would be \(-1\) since \(y=-x+6\) looks like \(y=-1\times x+6\)
general form is \(y=mx+b\) where \(m\) is the slope, so in this case \(m=-1\)

Answer 11

now we interpret the question as saying
"find the equation of the line with slope \(-1\) through the point \((-4,2)\) "
do you know how to do that?

Answer 12

not quite

Answer 13

ok this is what you need
\[y-y_1=m(x-x_1)\] this is the "point-slope" formula
you have a point, it is \((-4,2)\) this means \(x_1=-4,y_1=2\) and you have the slope \(m=-1\)
plug these numbers directly in to the formula
can you do that or would you like me to show it?

Answer 14

is it
y - 2 = -1 (x - 4)

Answer 15

oops, i was wrong
it is \(y-2=-1(x-(-4))\) or \[y-2=-(x+4)\]

Answer 16

I got it

Answer 17

then you still have one more job to do
you have to write it in the form \(y=mx+b\)
distribute the minus sign on the right and get
\[y-2=-x-4\] then add 2 to both sides to get
\[y=-x-2\]

Answer 18

these steps are always the same, so with a little practice you will be able to do them swiftly

Answer 19

Thank you, your a big help. I actually understand it more thn reading it in my lesson.

Answer 20

yw
try the next one'
question is
Write the equation of the line which passes through (2, –3) and is perpendicular to y = 4x + 7 in standard form.

slope of \(y=4x+7\) is 4, slope of perpendicular line is the "negative reciprocal" namely \(-\frac{1}{4}\)
this will require working with fractions

Answer 21

Okay

Answer 22

would it
(-1/4)(2) + x = x -5/2

Answer 23

you are missing a y somewhere

Answer 24

point is \((2,-3)\) use \(y-y_1=m(x-x_1)\) with \(x_1=2,y_1=-3, m=-\frac{1}{4}\)

Answer 25

y - (-3) = -1/4(x - 2)

Answer 26

yes, that is the correct first step

Answer 27

although you might want to go to \(y+3=-\frac{1}{4}(x-2)\) or else make that the second step

Answer 28

now distribute the \(-\frac{1}{4}\) on the right

Answer 29

y + 11/4 = (x - 2

Answer 30

just work on the right hand side, not the left

Answer 31

so would it be x = -7/4

Answer 32

\[y+3=-\frac{1}{4}(x-2)\]
\[y+3=-\frac{1}{4}x+(-\frac{1}{4})\times (-2)\]by the distributive property
then
\[y+3=-\frac{1}{4}x+\frac{1}{2}\]

Answer 33

Oh

Answer 34

then subtract \(3\) from both sides to get \(y\) by itself
you get
\[y=-\frac{1}{4}x+\frac{1}{2}-3\] or \[y=-\frac{1}{4}x-\frac{5}{2}\]

Answer 35

don't forget you need an \(x\) and a \(y\) in your answer
your answer needs to look like \(y=mx+b\)
this is not a "solve for \(x\)" problem, you do not want to end up with \(x=\text{some number}\)

Answer 36

will I do anything to it, if so am I adding 5/2 to each side

Answer 37

no it is good like it is
it looks just like \(y=mx+b\) where \(m=-\frac{1}{4}\) and \(b=-\frac{5}{2}\)

Answer 38

oooooooooooooh hold on

Answer 39

oh okay I got what your saying now

Answer 40

i am sorry, it says in "standard form"

Answer 41

lets start here
\[y+3=-\frac{1}{4}(x-2)\] and make it look like \(ax+by=c\)

Answer 42

multiply both sides by 4 to clear the fraction. you get
\[4y+12=-(x-2)\] or
\[4y+12=-x+2\] add \(x\) to both sides, gives
\[x+4y+12=2\] subtract \(12\) form both sides you get
\[x+4y=-10\] which looks just like \(ax+by=c\)

Answer 43

thank you this was less omplicated than the other lol