Question 34723This question is from textbook College algebra
: I need help. I don't know where to go from here. Okay, questions is; write and equations(a) in standard form and (b) in slope-intercept form.
Problem; through -(2, 0), Perpendicular to 8x-3y=7
This is what I came with and I don't know what to do after, if I even got this much right? y-0=3/8 (x-(-2), Y-0=3/8x+6/8...Thank you for your help.
This question is from textbook College algebra
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! straight line will take the form y=mx+c. We needs its gradient (m) and y-intercept (c).
Perpendicular to 8x-3y=7. OK, so first we need to write this like y=mx+c so we know its gradient...
8x-3y=7
8x-7 = 3y
or 3y = 8x-7
y = (8/3)x-7/3
Its gradient is 8/3. Our needed line, being perpendicular to this, will therefore have a gradient of -(3/8)... How do i know? invert the given gradient and change sign, basically to find the perpendicular one..eg for gradient of 3, perpendicular gradient is -1/3.
So, we have y=-(3/8)x + c. Now for c. Well we need to know a x and y pair, which we do: (2,0) or is it (-2,0)? I am picking the first. If the latter, then you will have to correct my working from here in.
0=-(3/8)(2) + c
0=-(6/8) + c
c = 6/8
--> equation is y=-(3/8)x + 6/8
This is slope/intercept form.
As for "standard form" i am not sure what your syllabus calls this, so i shall just write it with no fractions and no negatives:
y=-(3/8)x + 6/8
8y=-3x + 6
8y+3x = 6
Hopefully this is what you mean by standard form?
jon.
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