anonymous
  • anonymous
Help with these:/ I don't understand
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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hartnn
  • hartnn
can you find slope of any given line ?
hartnn
  • hartnn
like say , x-5y = -10 is a line. can you find slope of this line first ?

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anonymous
  • anonymous
Uh no not really... with a quick review I probably can
hartnn
  • hartnn
good, so i have x-5y =-10 i will try to isolate y and bring it in the form y=mx+c where m=slope so, x=5y-10 5y = x+10 y= (1/5)x +2 got this ?
anonymous
  • anonymous
Yeah I got it
hartnn
  • hartnn
cool, so can you find the slope of x -3y=9 line ? in same way
hartnn
  • hartnn
oh and i forgot to mention, y= (1/5)x +2 comparing with y=mx+c gives, m=slope = 1/5
anonymous
  • anonymous
x-37=9 -3y=x+9 Y=(1/3)x+3
anonymous
  • anonymous
So slope would be 1/3 I assume?
hartnn
  • hartnn
correct! :) now try to find slopes of other 3 lines also! and tell me, so that i will verify :)
anonymous
  • anonymous
4x+2y=5 2y=-4x+5 Y-(-1/2)x+2.5 ?
hartnn
  • hartnn
2y=-4x+5 is correct, but then divide both sides by 2 what u get ?
anonymous
  • anonymous
10x-5y=8 -5y=10x+8 Y=-2x+1.6?
hartnn
  • hartnn
10x-5y=8 -5y=10x+8 <<<<<
anonymous
  • anonymous
and you would get y=-2+2.5
anonymous
  • anonymous
Ahh okay -10 so the problem would come out with a positive 2x
hartnn
  • hartnn
yes, so the slope will be m= 2 , right ?
anonymous
  • anonymous
Yes
hartnn
  • hartnn
4x+2y=5 slope = -2 make a note of all slopes, we'll need them
hartnn
  • hartnn
and what about last line ?
anonymous
  • anonymous
4x+y=-1 Y=-4x-1
hartnn
  • hartnn
so, slope =?
anonymous
  • anonymous
-4
hartnn
  • hartnn
Good! now here's the rule : slope of parallel lines are equal! so, for the last one, slope of line parallel to 4x+y=-1 will also be m=-4 :) so, for last, the correct option will be `C) m=-4` got this ? in same way, can you match the 2nd last and 3rd last ?
anonymous
  • anonymous
Yeah I got it, Thanks a ton man. You need to be my teacher xD
hartnn
  • hartnn
haha, :) but we still require to do the 1st 2 !
hartnn
  • hartnn
another rule, `product of slopes of perpendicular lines = -1 ` so, for 1st one we had slope = 1/5 let slope of perpendicular line be m. so, \(\Large m\times (1/5)=-1\) find m from here :)
anonymous
  • anonymous
1st one is m=1/5? 2cnd one is m=1/3? It seems like these would be the answers but they are not possible to be selected.
anonymous
  • anonymous
Ahhhh okay
hartnn
  • hartnn
because perpendicular lines follow different rule :)
anonymous
  • anonymous
Well I still don't exactly get how it goes from (1/5) to -1...
hartnn
  • hartnn
product of slopes = -1 is the rule so, m (1/5) =-1 multiplying by 5, m = -5 got this ?
hartnn
  • hartnn
multiplying by 5 on both sides.
anonymous
  • anonymous
I'm confused on the product of slopes part, I get the multiplying on both sides. So like can you explain the product of slopes rule.. I tried looking it up on google but nothing I can comprehend really comes up.
hartnn
  • hartnn
if say line 1 has slope M1 and line 2 has slope M2 now if these 2 lines are perpendicular to each other, then M1*M2 =-1
hartnn
  • hartnn
another way to remember the same thing is, the slope of line is negative reciprocal of the perpendicular line, so, slope of line perpendicular to line with slope 1/5 will be - (1/(1/5)) = -5
hartnn
  • hartnn
so, thats your 1st one, m =-5 for 2nd one, we had slope = 1/3, so m *(1/3) = -1 so, m=-3 is the answer to 2nd one :)
anonymous
  • anonymous
Okay, so basically whenever there is a fraction just take (1/_) blank space number lol
hartnn
  • hartnn
you forgot the negative :P -1/ stuff
hartnn
  • hartnn
but this only applies to perpendicular lines
hartnn
  • hartnn
for parallel lines, slope are equal
anonymous
  • anonymous
Oh I see says the blind man
hartnn
  • hartnn
what who ?
hartnn
  • hartnn
lol
hartnn
  • hartnn
i hope you got the entire problem clearly ?
hartnn
  • hartnn
ask if any doubts anywhere
anonymous
  • anonymous
Yeah I get it now man thanks a ton
hartnn
  • hartnn
welcome ^_^

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