find an equation of the line having the slope and containg the given point m=3/4,(3,-2)
hes apparently in a exam @lgbasallote
hmm
we can write (y-(-2))=(3/4)(x-3) y=3x/4-1/4
let's give him the benefit of the doubt...just tell him HOW to solve it..
thanks drewnj
find an equation of the line containing the points (4,1) and (12,8)
How about you use the method shown above^^ by @AndrewNJ ?
ok
Just find your slope and then use the method shown.
first u have to find out the slpoe k k=(y1-y2)/(x1-x2)=(8-1)/(12-4)=7/8 then substitution the slope k and one point [(4,1) or (12,8) ]in to a basic quadratic equation y=kx+b if i use (4,1) then 1=(7/8)*4+b we can work out b=-5/2 so y=(7/8)x-5/2
u are right! the work you done can have the slope can substitution it and a point into y=kx+b to have the b
the second 'can' should be 'and' -.-
so -7=(5/7)*0+b right
you reverse x and y it should be 0=(5/7)*(-7)+b remember y=kx+b
:)
so it should read y=(5/7)x-1
0=(5/7)*(-7)+b 0=-5+b b=5
so it should read y=(5/7)x+5 right
yes!
ok i am getting it
what's your mean for it ?
write an equation of the line containing the given point and perpendicular to the given line
rewrite 2x+9y=4 to y=(-2/9)x+4/9 and (-2/9) is the slope
no it just work out the slope u still need to sub the slope and the point (0,7) into y=kx+b to find b
couldnt you also write it as 2(0)+9(7)=4
oh im sorry the upper one is perpendicular, i thought it was Parallel i redo it again
rewrite 2x+9y=4 to y=(-2/9)x+4/9 and the slope of that line is -2/9 so the slope perpendicular to that line is k=1/(-2/9)=-9/2 and then substitute k=-9/2 and point (0,7) into y=kx+b to find b
so the equation should read 0=-9/2*7+b
this time the point is (0,7) x=0, y=7. so 7=(-9/2)*0+b
oh yea thats right
lol
so next line should read 7=0+b then y=-7+b
so we work out b=7 and FINALLY substitute k=-9/2 and b=7 into y=kx+b the final equation is y=(-9/2)x+7
ok whatever you say
you a smart dude
:)
what grade you in
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