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)
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?
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)
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
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?
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
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.
try to make progress if you post your work someone (me? if I'm here) will check it.
okay. for 4 and 5 what do you mean by plot lines?
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.
oh okay, I see.
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?
what two points does it go through. I'm not sure which line you mean
one second. Ill show the updated picture
I cut the triangle in half to make it how you said it would be easier.
the line that runs alongside the left side of the triangle
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
okay, so where would I go from there
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
okay, so would i move it up just a little?
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)
Okay. The (4,8) is the lower left part of the rectangle..
ok. after you fix the line we can do Q4
Okay:)
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
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 ?
so would it just be y=(1/-3)x+2?
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)
okay, so would the (3/2) be the same then?
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 ?
yes, i already did 1,2, and 3
You have to choose 2 locations that make a line and are either positive or negative. I
what did you choose?
i chose (2,5) (12,19) for positive (21,5) (12,19) for negative.
so for Q4, would it still be y=(3/2) +b
but we still have to find b
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)
okay
and change (21,5) (12,19) for negative. to (21,5), (19, 8) (same reason..a nicer slope)
make both of them negative?
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)
Ohhh okay haha
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 ?
would i solve for b
yes. can you do that?
I mean do i isolate it
yes
okay so i got (-3/1)-(3/2)
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
okay so it would be -4.5?
yes, or -9/2 (same thing)
so the answer for Q4 is y = (3/2) x - 9/2
y=(3/2)*1+(-9/2)?
ok you already said that.
sorry haha
right, sorry
that was disgusting
wow. okay. im sorry you guys had to go through that.
i blocked them
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 ?
(-3/2)?
since theyre reciprocals?
we want "flip and negate" you negated but did not flip
(-2/3)
yes so we have y = (-2/3) x + b and they want this to go through point (-2,7)
to find b, replace x and y with those numbers
okay so it would be 7=(-2/3)*-2+b
yes
then solve for b again right
yes
7/-2- (-2/3)=b
sorry i didnt put parenthesis on the first pair
that looks a bit peculiar. start with 7=(-2/3)*-2+b can you do -2 * -2/3 ?
(-4/6)?
(4,6)
sorry forgot a negative times a positive for a second. haha
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
4/3?
yes. so you have 7 = 4/3 + b now add -4/3 to both sides
(7/-2)+(4/3)=b?
wait whoops
hold on, sorry
7+4/3? i was looking at my old equation
close. but added +4/3 we want -4/3 (because on the right side we want 4/3 - 4/3 which cancels)
im sorry, thank you for being so patient with me. math really isnt my stronest subject... at all.
okay, gotcha
so just 7+(-4/3)
b= 7 - 4/3 change 7/1 to 21/3 (multiply top and bottom by 3)
and yes you can write it 7 + (-4/3)
so 25/3?
21/3+4/3=25/3
if you did 21/3 + 4/3 we have 21/3 - 4/3
oh right ints negative
17/3
and the equation is ?
Join our real-time social learning platform and learn together with your friends!