Algebra HELP ASAP?

Question

I have learned to wait until the last min to do my homework and i am totally lost.

Find the equation, in standard form, with all inter coefficients, of the line perpendicular to x+ 3y =6 and passing through (-3, 5). Show all work

Answer 1

..............................................................
...............ALGEBRA.................................
http://www.algebra-helper.com/quadratic-equations-1.htm
This link provides a multitude of sub links all with written equations
adding fractions
brackets
comparing fractions
complex fractions
decimals
difference squares
dividing fractions

Algebra is not my forte' but I can give you the above link for future help.
.....................MATH.............................
http://mathforum.org/library/drmath/view/54242.html

Answer 2

x+3y=6 is the same as 3y= -x+6

divide the equation by 3 to give y= -x/3 +2

therefore the gradient of that line is minus a third (-1/3) and when lines are perpendicular the gradients will be the negative percipricle or something or other. for example if the gradient is "a" of one line, then the other line will be "-1/a". so the gradient in question will be simply 3.

then you follow the basic y=mx+c pattern to find the equation. substitute in the points you are given and the gradient to find c.

5=(3x-3) +c
5= 9 +c
-4=c

then you simply write the y=mx=c equation filling in the m and c.

answer: y=3x-4

i hope i helped you, i dont know what you mean by inter coefficients but i answered it as best i could!