WRITE IN STANDARD FORM the equation of the line goin thru points (7,5) & (-4,9) use (y2-y1),(x2-x1) formula
can u find the slope ?
yes its - 4/11
y-y1 = m (x-x1 ) u have m now and a point too.. plug in this formula u will get ur answer and standard form look like this y= mx + c
i already did do that bro watch y-5= -4/11(x-7) y-5=-4/11x + 28/11 (4)y-5=-4/11+28/11(4) 4y-20=-1x+7 add 1x to both sides 1x+4y-20=7 add 20 to both sides 1x+4y=27 am i right?
what u did in 3rd step/?
and beside its wrong..
the equation u ahve found does not have a slope u have found above.
to make it into standard form u have to multipy both sides of the equation by the LCD and if its wrong help me solve it then?
y-5= -4/11(x-7) ok this is the equation .. now we have to multiply and reaarange it.. first.. can we multiply both sides with 11 so we can get rid of denominator .. agreed?
if i knew i wouldnt really be askin the question -.- but yes i guess so...keep goin?
we are just multiplying.. thats all so after multiplying 11 we will get 11(y-5) = -4 (x-7) 11y-55 = -4x +28... can u do it from here?
thats exactly what i got when i worked it the second time!!! awesome but yes hold up i ended up with 4x+11y=27
hmm no.. there shouldnt be 27 4x+11y = 28+55
thats not in STANDARD FORM tho
so its 4x+11y=83 ?
yes i know .. standard form is something like this y = mx +c so we have to reaarange it again 4x+11y=83 11y = -4x +83 y =-4/11x +83/11 this is ur standard form.
nah standard form is Ax+By= C (no fractions coefficeint of X is positive A > 0 -
lol oppss ya that was slope interception form.... 4x+11y=83 so this would be it
told u :D
lol. ya my bad..
what about same problem but with these points (6,5) (-1,9)
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