anonymous
  • anonymous
The table shows the proof of the relationship between the slopes of two perpendicular lines. What is the missing statement in step 2?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
1. is similar to . given 2. property of similar triangles 3. = property of proportion 4. slope of = slope of = definition of slope 5. slope of × slope of = multiplying the slopes 6. (This statement is intentionally left blank.) Substitution Property of Equality 7. slope of × slope of = -1 simplifying the right side
anonymous
  • anonymous
couldnt get all of it but tried my best
jim_thompson5910
  • jim_thompson5910
a screenshot is best because we can see the full problem (even the missing symbols)

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anonymous
  • anonymous
1. AC←→⊥CD←→ (view diagram) ΔABC is similar to ΔCED. given 2. (This statement is intentionally left blank.) property of similar triangles 3. ABBC = ECDE property of proportion 4. slope of AC←→ = -ABBC slope of DC←→ = DEAC definition of slope 5. slope of AC←→ × slope of DC←→ = -ABBC×DEEC multiplying the slopes 6. Substitution Property of Equality 7. slope of AC←→ × slope of DC←→ = -1 simplifying the right side
anonymous
  • anonymous
sorry here it is
jim_thompson5910
  • jim_thompson5910
well step 3 has AB*BC = EC*DE, or so it seems the step before that would be some proportion and cross multiplying would lead to step 3
jim_thompson5910
  • jim_thompson5910
does that make sense?
anonymous
  • anonymous
not really but u can explain it in detail. mabye ill get it
jim_thompson5910
  • jim_thompson5910
let's say you had a/b = c/d
jim_thompson5910
  • jim_thompson5910
if you cross multiplied, you'd get a*d = b*c agreed?
anonymous
  • anonymous
yea
jim_thompson5910
  • jim_thompson5910
so what proportion could be set up to lead to AB*BC = EC*DE ?
anonymous
  • anonymous
they seem to look like this but in fraction AB/BC;. something like that.
anonymous
  • anonymous
AB/DE, and EC/DE
jim_thompson5910
  • jim_thompson5910
more like AB/EC = DE/BC. There are other possibilities |dw:1436929680423:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1436929728958:dw|
anonymous
  • anonymous
oops sorry. thats not on my answer choices tho
anonymous
  • anonymous
the closest choice has a fraction DE/EC. would that work?
jim_thompson5910
  • jim_thompson5910
well as long as they cross multiply to AB*BC = EC*DE, then it will work
jim_thompson5910
  • jim_thompson5910
there are more possibilities than what I drew up there (maybe 4 possibilities?)
anonymous
  • anonymous
ok hold on
anonymous
  • anonymous
AB/EC=BC/DE AB/BC=DE/EC AB=DE BC=EC AB/BC= -EC/DE
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
choice C looks funky...there are no fractions there?
jim_thompson5910
  • jim_thompson5910
try posting a screenshot if you can
anonymous
  • anonymous
there are no fractions in c, and sorry i cant post a screenshot.
jim_thompson5910
  • jim_thompson5910
ok let me think
jim_thompson5910
  • jim_thompson5910
in line 3, where it says `3. ABBC = ECDE property of proportion` there are no fractions here either?
anonymous
  • anonymous
actually yea those are fractions
anonymous
  • anonymous
its really hard to paste those kinds of things on here.
jim_thompson5910
  • jim_thompson5910
so line 3 actually says this right? AB/BC = EC/DE
anonymous
  • anonymous
yes yes
jim_thompson5910
  • jim_thompson5910
|dw:1436930874242:dw|
jim_thompson5910
  • jim_thompson5910
you can multiply both sides by BC and divide both sides by EC that will make BC and EC swap |dw:1436930925056:dw|
jim_thompson5910
  • jim_thompson5910
giving |dw:1436930949685:dw|
anonymous
  • anonymous
omg thanks so much!
anonymous
  • anonymous
can you help me with one more? its a mastery, and i didnt really understand this lesson
jim_thompson5910
  • jim_thompson5910
sure
anonymous
  • anonymous
AB and BC form a right angle at point B. If A = (-3, -1) and B = (4, 4), what is the equation of BC
jim_thompson5910
  • jim_thompson5910
first find the equation of the line through A and B
anonymous
  • anonymous
slope?
anonymous
  • anonymous
5/7 is the slope
jim_thompson5910
  • jim_thompson5910
use that slope to find the equation of the line through those two points
jim_thompson5910
  • jim_thompson5910
you can use y = mx+b
anonymous
  • anonymous
i think its y=5/7x+8/7
jim_thompson5910
  • jim_thompson5910
yep
jim_thompson5910
  • jim_thompson5910
|dw:1436931657035:dw|
jim_thompson5910
  • jim_thompson5910
AB and BC form a right angle, so |dw:1436931712083:dw|
jim_thompson5910
  • jim_thompson5910
what is the slope of BC ?
anonymous
  • anonymous
x + 3y = 16 2x + y = 12 -7x − 5y = -48 7x − 5y = 48
anonymous
  • anonymous
these are the answer choices btw. and the slope for BC is -7/5 right?
jim_thompson5910
  • jim_thompson5910
what is the equation of the line going through B and has slope of -7/5
jim_thompson5910
  • jim_thompson5910
use y = mx+b

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