Please help!
I have an assignment that I have been trying to do but i'm a bit stuck. I have done the first questions but need help on the last ones. Help would be much appreciated. (i will put pictures below)

- tw101

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- tw101

##### 1 Attachment

- tw101

This is what I have. And I have already done questions 1,2,and 3, but only got halfway on 6&7. I drew the triangle and rectangle but I'm not sure how to get the length of the sides.Im also not sure how to do 4&5. I know that for parallel the slopes are the same and for perpendicular they're opposite reciprocals, but do I have to draw two new lines to accommodate the given points?

##### 1 Attachment

- phi

to do Q4, we need to know the equation of the line for Q2 (for example)
a parallel line will have the same slope
I don't know what slope you got for Q2, but pretend it was 2.
then the equation for Q4 would be
y = 2x + b
(same slope, but we need to find b)
they say this line goes through the point (1,-3) (i.e. when x is 1, y is -3)
so use that info to get
-3 = 2*1 + b
or
-3 = 2+ b
to find b, add -2 to both sides and simplify
but remember we need the slope for Q2 (which is probably not 2)

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- phi

for Q5
a perpendicular slope to slope "m" is -1/m
for example if m is 2/3 then the perpendicular slope is -3/2 (flip and negate)
or: -1/2 and 2
or 2 and -1/2
or 3 and -1/3
say you choose the line from Q2. get its slope and "flip it"
then multiply by -1
after you get the slope, it is the same problem as Q4... you have to find b

- phi

for Q6, I would pick a "simple triangle" (a right triangle with legs that go straight across and straight up/down) we don't have to, but it is easier.
the length of the straight across line is easy: count the number of squares
(or subtract the x values)
for a slanted line you have to use the distance formula
do you have that in your notes?

- phi

I don't know if you have to plot the lines for Q4 or Q5
but it can't hurt. once you have the equation, find two points on the new line and plot them... then connect the dots

- tw101

THank you so much for helping me! sorry i didnt respond right away! I didnt realize that someone was actually looking at my question. I had given up haha.

- phi

try to make progress
if you post your work someone (me? if I'm here) will check it.

- tw101

okay. for 4 and 5 what do you mean by plot lines?

- phi

you wrote
*** but do I have to draw two new lines to accommodate the given points?***
which I assumed meant: do I have to plot the new lines which will be parallel or perpendicular to the original line?
and I thought it would not hurt.

- tw101

oh okay, I see.

- tw101

I was going to do the line I drew over the left side of the triangle since its goes through the points it gave me. would that work?

- phi

what two points does it go through. I'm not sure which line you mean

- tw101

one second. Ill show the updated picture

- tw101

I cut the triangle in half to make it how you said it would be easier.

- tw101

##### 1 Attachment

- tw101

the line that runs alongside the left side of the triangle

- phi

ok, but first we have to fix that line.
it goes through (2,5) and (4,8)
its slope is change in y divided by change in x
(8-5)/(4-2) or 3/2
we can say
y = (3/2)x + b
now use (2,5):
5 = (3/2)*2 + b
5= 3+b
b= 2
and the equation is y= (3/2)x + 2

- tw101

okay, so where would I go from there

- phi

notice, when y= 19 (top of the outer box)
19= (3/2)x + 2
add -2 to both sides
17 = 3/2 x
multiply both sides by 2/3
34/3 = x
or x = 11.333 (11 and 1/3)
not x=12 as you have. in other words your line on the graph is a little off

- tw101

okay, so would i move it up just a little?

- phi

not up. I would move the point at (12,19) to (11 1/3 , 19)
you will have to estimate 1/3 of box over from 11
then redraw the line to go through the 3 points
(2,5), (4,8) and (11 1/3, 19)

- tw101

Okay. The (4,8) is the lower left part of the rectangle..

- phi

ok. after you fix the line we can do Q4

- tw101

Okay:)

- phi

btw, y= (3/2) x + 2
will be an answer for Q2 . a line with positive slope
and you have the work up above on how we found the equation

- phi

Q4 ... the question is a bit muddled but I assume it is the new line that goes through (1,-3)
and they want it parallel to
y= (3/2) x + 2
parallel means the same slope.
so write the new equation. can you do that ?

- tw101

so would it just be y=(1/-3)x+2?

- phi

start with
y = m x + b (the "skeleton" that we fill in)
we want this line to be parallel to y= (3/2)x + 2
what slope should we use for m ? (parallel means *same* slope)

- tw101

okay, so would the (3/2) be the same then?

- phi

Q1 is making the graph. you mean Q2 and Q3?
for the "old line" we can use the line you used for Q2
what was it ?

- tw101

yes, i already did 1,2, and 3

- tw101

You have to choose 2 locations that make a line and are either positive or negative. I

- phi

what did you choose?

- tw101

i chose (2,5) (12,19) for positive
(21,5) (12,19) for negative.

- tw101

so for Q4, would it still be y=(3/2) +b

- tw101

but we still have to find b

- phi

if you change to (2,5), (4,8) we will get 3/2 for the slope (a much nicer number than if we use (12,19)

- tw101

okay

- phi

and change (21,5) (12,19) for negative. to (21,5), (19, 8)
(same reason..a nicer slope)

- tw101

make both of them negative?

- phi

no. for Q2 use the points (2,5) and (4,8) (slope will be 3/2)
for Q3 use the points (21,5), (19,8) (slope will be -3/2)

- tw101

Ohhh okay haha

- phi

I hope that gets straightened out.
but for Q4, we want a line parallel to the line from Q1 which has a slope of 3/2
we start with
y= mx + b
we replace m with 3/2
y= (3/2) x + b
we want this new line to go through (1,-3)
replace x and y with those numbers
-3 = (3/2)*1 + b
can you find b ?

- tw101

would i solve for b

- phi

yes. can you do that?

- tw101

I mean do i isolate it

- tw101

yes

- tw101

okay so i got (-3/1)-(3/2)

- phi

yes, but people would combine that
either put -3 over the common denominator of 2 (i.e. write it as -6/2 )
or change 3/2 to 1.5 and figure out -3 - 1.5

- tw101

okay so it would be -4.5?

- phi

yes, or -9/2 (same thing)

- phi

so the answer for Q4 is y = (3/2) x - 9/2

- tw101

y=(3/2)*1+(-9/2)?

- tw101

ok you already said that.

- tw101

sorry haha

- tw101

right, sorry

- tw101

that was disgusting

- tw101

wow. okay. im sorry you guys had to go through that.

- tw101

i blocked them

- phi

ready for Q5?
let's use the same line with slope (3/2)
and we want a perpendicular line
y= mx + b
any idea what we use for m ?

- tw101

(-3/2)?

- tw101

since theyre reciprocals?

- phi

we want "flip and negate"
you negated but did not flip

- tw101

(-2/3)

- phi

yes
so we have
y = (-2/3) x + b
and they want this to go through point (-2,7)

- phi

to find b, replace x and y with those numbers

- tw101

okay so it would be 7=(-2/3)*-2+b

- phi

yes

- tw101

then solve for b again right

- phi

yes

- tw101

7/-2- (-2/3)=b

- tw101

sorry i didnt put parenthesis on the first pair

- phi

that looks a bit peculiar.
start with
7=(-2/3)*-2+b
can you do -2 * -2/3
?

- tw101

(-4/6)?

- tw101

(4,6)

- tw101

sorry forgot a negative times a positive for a second. haha

- phi

it is
\[ -2 \cdot \frac{-2}{3} = \frac{-2}{1} \cdot \frac{-2}{3}\]
when you multiply fractions you multiply top * top
and bottom * bottom

- tw101

4/3?

- phi

yes.
so you have
7 = 4/3 + b
now add -4/3 to both sides

- tw101

(7/-2)+(4/3)=b?

- tw101

wait whoops

- tw101

hold on, sorry

- tw101

7+4/3? i was looking at my old equation

- phi

close. but added +4/3
we want -4/3 (because on the right side we want 4/3 - 4/3 which cancels)

- tw101

im sorry, thank you for being so patient with me. math really isnt my stronest subject... at all.

- tw101

okay, gotcha

- tw101

so just 7+(-4/3)

- phi

b= 7 - 4/3
change 7/1 to 21/3 (multiply top and bottom by 3)

- phi

and yes you can write it 7 + (-4/3)

- tw101

so 25/3?

- tw101

21/3+4/3=25/3

- phi

if you did 21/3 + 4/3
we have 21/3 - 4/3

- tw101

oh right ints negative

- tw101

17/3

- phi

and the equation is ?

- tw101

y=(-2/3)x+17/3?

- phi

yes

- tw101

okay great

- phi

for Q6, did you change the top point from (12,19) to (11 1/3 , 19) ?

- tw101

of the triangle?

- phi

yes (otherwise you will get marked off (if they notice!) because (12, 19) is not really on the line from (2,5) to (4,8)

- tw101

yep

- tw101

just draw a side straight down from that point now

- phi

ok, now figure out the slope of the straight up/down line
can you do that ?

- tw101

i think so

- tw101

jsut a minute:)

- tw101

okay so the two points are (11 1/3,19) and (11 1/3,4)
so then i did the y1-y2/x1-x2 thing and got (0,15) not sure what to do now. would i substitute that into the equation?
y=mx+b

- phi

they don't want the equation of the line. they want its slope
change in y divided by change in x
notice you will get (19-4)/ (11.333- 11.333) = 15/0
and divide by 0 is not allowed. we say the slope of a vertical line is undefined.

- phi

they also want the length of that side. it is 19-4 or 15

- tw101

okay so slope:undefined length:15

- phi

yes. now do the bottom side

- tw101

ok

- phi

though in your picture it looks like you only go down to y=5 (not y=4)

- phi

I would try to use the point (2,5) in this triangle because we know the slope of the slant side will be 3/2 (we did that work up above)

- tw101

youre right. 14 then

- tw101

sorry my internet is being really weird

- phi

yes. and change the number where you show your work.

- tw101

okay so for the bottom side i got (11 1/3,5) (2,5) then when subtracted i got -1/7 1/3

- phi

ok, now you can do the bottom line
(2,5) to (11 1/3, 5)
change in y is (5-5)
change in x is 11 1/3 - 2 = 9 1/3
and 0/ 9.333 is 0

- tw101

ugh no that is not what i got

- phi

if you have a calculator, it will help

- tw101

sorry my computer is freezing up

- phi

the length will be 9 1/3
(9 and 1/3)
slope will be 0

- phi

as a check, you can count how long it is. but they want you to "show the work"
so you should also show 11 1/3 - 2 = 9 1/3

- tw101

would the slope be 0?

- tw101

oh okay you said that haha sorry, again , slow computer

- phi

yes , slope is 0 (horizontal lines have slope 0)

- phi

now the slant side
(2,5) to (11 1/3, 19)
first find change in y and change in x

- phi

hint we did this already (see length of the vertical and horizontal sides)

- tw101

yes, sorry i had to reset my connection

- phi

you should get 14 and 9 1/3

- tw101

yes, i did. im so sorry idk what going on with my wifi

- tw101

change of x 19-5 change of y 11 1/3-2

- tw101

14 and 9 1/3 yes.

- phi

ok on your numbers, but x goes sideways... change in x is 9 1/3
the slope will be 14 / (9 1/3)
to do that, write 9 1/3 as an improper fraction 28/3
when you divide by a fraction, you can "flip it" and multiply
so 14 * 3/28 which is 3/2

- tw101

ok

- phi

slope is 3/2
length we use the distance formula
\[ \sqrt{ (y_2-y_1)^2 + (x_2-x_1)^2 } \]
which is
\[ \sqrt{ 14^2 + 9.3333^2} = \sqrt{196+87.11}= \sqrt{283.11}= 16.8\]

- phi

because the triangle has a right angle formed by the vertical and horizontal lines, it is a right triangle

- tw101

oh okay i see.

- phi

for Q7 I would use the "mall" as the quadrilateral
the slopes will be 0 for the top and bottom (horizontal lines)
and undefined for the sides
the lengths are easy (I hope... count how long each side is)
because the shape has 4 90 degree angles and the opposite sides are equal
it is a rectangle.

- phi

I have to go.

- tw101

9 ad 15

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