Question 34723
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.