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Mathematics 9 Online
OpenStudy (anonymous):

I really need help at understanding perpendicular lines. I kinda understand paralel lines, but not so well. My teacher quickly brushed over it and we have a test tommorow but I don't know anything about it :( I need to find perpendicular lines without a graph and she didnt teach us it.

jimthompson5910 (jim_thompson5910):

so you're given equations?

OpenStudy (anonymous):

Yeah. She said inn the test she will give us in y=mx+b and (x,y).

jimthompson5910 (jim_thompson5910):

do you have a sample problem?

OpenStudy (anonymous):

no

jimthompson5910 (jim_thompson5910):

do you want me to make one up?

OpenStudy (anonymous):

If you can that would be great :P

jimthompson5910 (jim_thompson5910):

What is the equation of the perpendicular line to y = 3x+7 and this perpendicular line goes through (1,5)

OpenStudy (anonymous):

Im not quite sure how to find it :(

jimthompson5910 (jim_thompson5910):

what is the slope of y = 3x+7

OpenStudy (anonymous):

3

jimthompson5910 (jim_thompson5910):

you can think of 3 as the fraction 3/1 since 3/1 = 3

OpenStudy (anonymous):

ok

jimthompson5910 (jim_thompson5910):

take the original slope and flip the fraction 3/1 --> 1/3 then flip the sign +1/3 --> -1/3

jimthompson5910 (jim_thompson5910):

original slope: 3/1 perpendicular slope: -1/3

OpenStudy (anonymous):

oh

OpenStudy (anonymous):

that is it?

jimthompson5910 (jim_thompson5910):

we know two things perpendicular slope: m = -1/3 the perpendicular line goes through the point (x,y) = (1,5)

jimthompson5910 (jim_thompson5910):

so you take m = -1/3, x = 1, y = 5 and plug them into y = mx+b

jimthompson5910 (jim_thompson5910):

and then solve for b

OpenStudy (anonymous):

so it would be 5=1/3(1/1)+b before simpylfying?

jimthompson5910 (jim_thompson5910):

if you mean \[\Large 5 = \frac{1}{3}\left(\frac{1}{1}\right) + b\] then yes. Now you solve for b

jimthompson5910 (jim_thompson5910):

oops

jimthompson5910 (jim_thompson5910):

the 1/3 should be negative

jimthompson5910 (jim_thompson5910):

\[\Large 5 = -\frac{1}{3}\left(\frac{1}{1}\right) + b\]

OpenStudy (anonymous):

ok so it would be like 5= -1/3+b then divide 5 by -1/3 which would be -15

OpenStudy (anonymous):

so b=-15?

jimthompson5910 (jim_thompson5910):

you add 1/3 to both sides to move that 1/3 over, which isolates b

OpenStudy (anonymous):

oh so it'll actually be\[5 \frac{ 1 }{ 3 }=b\]

jimthompson5910 (jim_thompson5910):

correct since \[\Large 5 + \frac{1}{3} = 5 \frac{1}{3}\]

OpenStudy (anonymous):

oh cool!

OpenStudy (anonymous):

so then what do we do after that?

jimthompson5910 (jim_thompson5910):

m = -1/3 and b = 5 & 1/3 so you have enough info to know the full perpendicular equation in y = mx+b form

OpenStudy (anonymous):

so y= -1/3+5 and 1/3 is the perpendicular line?

jimthompson5910 (jim_thompson5910):

more like \[\Large y = -\frac{1}{3}x + 5 \frac{1}{3}\]

OpenStudy (anonymous):

right i forgot the x

jimthompson5910 (jim_thompson5910):

since 5 & 1/3 = 16/3, we can also write it as \[\Large y = -\frac{1}{3}x + \frac{16}{3}\]

OpenStudy (anonymous):

you are a life saver :D

OpenStudy (anonymous):

thank you so much!

jimthompson5910 (jim_thompson5910):

you're welcome here is more practice http://www.mathworksheets4kids.com/straight-lines/perpendicular-lines-equations.pdf it offers practice problems & answers as well (just not steps)

OpenStudy (anonymous):

thank you :)))

OpenStudy (anonymous):

i wish my teacher explained like you xD lol

jimthompson5910 (jim_thompson5910):

no problem

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