Find the equation of the line through (4,3) perpendicular to the line 3x+y=7
so this is going to be similiar to what we did, but for a perpendicular line, the slope is the opposite reciprocal
it always changes on a perpendicular line?
i'm not sure what you mean
if you have two lines that are parallel, they have the same slope if you have two lines that are perpendicular, than their slopes are opposite reciprocals
like the slope is the opposite reciprocal?
yeah, for any two lines to be perpendicular, it means their slope are opposite reciprocals
ooo ok got it
you can think of it visually too if you have two lines that are parallel, it means that they never touch, right? so that means that they're at exactly the same angle, like so: |dw:1337182568814:dw| these two lines will always have the same slope |dw:1337182598672:dw| these lines are perpendicular, meaning they're the exact opposite of each other, and they intersect at a 90degree angle so if the slope is 5 for one, the slope for the other will be - 1/5
ooo ok yea it makes sense
so using what you learned on the previous problem, can you tell me what the equation is?
ok so i would put 3x+y=7 then it would be y=3x+7 right?
keep in mind that you're not just moving it over, you're actually subtracting it from both sides
o right so it would be y=-3x+7 right?
yup!
so what's the slope of your perpendicular line?
ok so now i do the y-y1=m (x-x2) right?
not quite yet, you don't have your correct slope yet. remember you're looking for the slope of a line that is perpendicular to the line you just gave me
ooo so i multiply -1 to -3 and +7?
na you don't even need to do any manipulation to it, just know that the reciprocal of something is the opposite of it so like 2/3 = 3/2 5/4 = 4/5 1/2 = 2/1 5 = 1/5 and if you're taking the opposite, than it's just going to be negative, if your number is positive, and positive if your number is negative so the opposite reiprocal of -9 will be 1/9 and the opposite reciprocal of 7/6 will be -6/7 make sense? so for the line you just gave me, what's its slope and what is its opposite reciprocal?
so it would be 1/3?
yup :o
yay ok so it would be y=1/3+7 right?
well, you just figured out the slope, now as you said before you have to plug in the two points (4,3) into y-y1 = m (x - x1) you're almost there!!
ok but when u said 5=1/5 would it be negative too?
i was just giving examples of reciprocals, without adding in the opposite sign yet the reciprocal of 5 = 1/5, the opposite reciprocal of 5 = -1/5 make sense?
ook but was i right?
which part?
thats 5= 1-/5?
it depends on what you're looking for.. if you're trying to find the opposite reciprocal, then yes
the word opposite means that it changes signs (negative to positive, positive to negative) the word reciprocal just means that it flips
oo ok thnx ok going on to the other problem plug in the two points
excellent ^.^
ok i hope im right lol i did y-3=1/3(x-4) right?
then i did 3(y-3)=x-4
the first part looks right, but what happened with the y in the second part?
remember, you want to get the y by itself, so you want to keep that 1/3 over on the right
are you familiar with distribution?
yea i am but i thought u leave x were 4 is and 3 on the other side
what you want to do is distribute the 1/3 to the (x-4)
yes
just ignore the left side for now, and try to do the distribution
so u multiply 1/3 with -4?
yup
and the x as well
you're literally just distributing the 1/3 into the (x-4)
u lost me lol
you said you're familiar with distribution right? in this case, you're distributing the 1/3 into the (x-4) so, you're multiplying both terms by 1/3 like so: (x(1/3) - 4(1/3))
o yea so it would be1/3x but the other 1 i dont know wat it would be
if you're multplying 4 by 1/3 it's the same thing as multiplying 4/1 x 1/3, does that make it easier?
\[\left(\begin{matrix}4 \\ 1\end{matrix}\right) x \left(\begin{matrix}1 \\ 3\end{matrix}\right)\]
ok so it would be 4/3?
yup ^.^
so now, we can look back over at the left side y-3 = 1/3x+4/3
now, you want to get the y alone, so what should you do?
ok i would move that negative and put it positve and put it on the other side and add 4/3?
yup!!! you got it ^.^
just remember that you can't just add it as it is though, you're dealing with 3/1 + 4/3, so you have to get a common denominator
ok but how would i add 4/3 to 3?
\[\left(\begin{matrix}4 \\ 3\end{matrix}\right) + \left(\begin{matrix}3 \\ ?\end{matrix}\right)\]
oops that ? should be a 1 lol
lol i was gonna say would it be a 1?
basically you want to make it so that the two bottom numbers match, so you want it to be something /3 .. whatever you did to the bottom to get that 3, you have to do to the top as well
\[\left(\begin{matrix}3 \\ 1\end{matrix}\right) = \left(\begin{matrix}9 \\ 3\end{matrix}\right)\], right?
can you see how i got that?
no i dont im srry
so i want to make the bottom match to 4/3 right? i need to do soething to 3/1 so that it has the same bottom ?/3 i multiplied the bottom by 3, to get ?/3 and whatever i did to the bottom, i have to do to the top as well so 3x3=9 /3
it's kind of hard to explain over text lol
its ok i get it alittle bit
but thank you for trying to explain
when you're adding fractions, the denominators have to be equal.. so if they're not, you have to do something to the bottoms to make them equal,a nd whatever you do to the bottom, you have to do to the top as well
http://www.mathsisfun.com/fractions_addition.html that explains it pretty well
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