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

Algebra ->  Linear-equations -> SOLUTION: Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3, 5).       Log On


   

Question 133144: Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3, 5).

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
First convert the standard equation x%2B3y=6 into slope intercept form

Solved by pluggable solver: Converting Linear Equations in Standard form to Slope-Intercept Form (and vice versa)
Convert from standard form (Ax+By = C) to slope-intercept form (y = mx+b)


1x%2B3y=6 Start with the given equation


1x%2B3y-1x=6-1x Subtract 1x from both sides


3y=-1x%2B6 Simplify


%283y%29%2F%283%29=%28-1x%2B6%29%2F%283%29 Divide both sides by 3 to isolate y


y+=+%28-1x%29%2F%283%29%2B%286%29%2F%283%29 Break up the fraction on the right hand side


y+=+%28-1%2F3%29x%2B2 Reduce and simplify


The original equation 1x%2B3y=6 (standard form) is equivalent to y+=+%28-1%2F3%29x%2B2 (slope-intercept form)


The equation y+=+%28-1%2F3%29x%2B2 is in the form y=mx%2Bb where m=-1%2F3 is the slope and b=2 is the y intercept.







Now let's find the equation of the line that is perpendicular to y=%28-1%2F3%29x%2B2 which goes through (-3,5)

Solved by pluggable solver: Finding the Equation of a Line Parallel or Perpendicular to a Given Line


Remember, any two perpendicular lines are negative reciprocals of each other. So if you're given the slope of -1%2F3, you can find the perpendicular slope by this formula:

m%5Bp%5D=-1%2Fm where m%5Bp%5D is the perpendicular slope


m%5Bp%5D=-1%2F%28-1%2F3%29 So plug in the given slope to find the perpendicular slope



m%5Bp%5D=%28-1%2F1%29%283%2F-1%29 When you divide fractions, you multiply the first fraction (which is really 1%2F1) by the reciprocal of the second



m%5Bp%5D=3%2F1 Multiply the fractions.


So the perpendicular slope is 3



So now we know the slope of the unknown line is 3 (its the negative reciprocal of -1%2F3 from the line y=%28-1%2F3%29%2Ax%2B2). Also since the unknown line goes through (-3,5), we can find the equation by plugging in this info into the point-slope formula

Point-Slope Formula:

y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope and (x%5B1%5D,y%5B1%5D) is the given point



y-5=3%2A%28x%2B3%29 Plug in m=3, x%5B1%5D=-3, and y%5B1%5D=5



y-5=3%2Ax-%283%29%28-3%29 Distribute 3



y-5=3%2Ax%2B9 Multiply



y=3%2Ax%2B9%2B5Add 5 to both sides to isolate y

y=3%2Ax%2B14 Combine like terms

So the equation of the line that is perpendicular to y=%28-1%2F3%29%2Ax%2B2 and goes through (-3,5) is y=3%2Ax%2B14


So here are the graphs of the equations y=%28-1%2F3%29%2Ax%2B2 and y=3%2Ax%2B14




graph of the given equation y=%28-1%2F3%29%2Ax%2B2 (red) and graph of the line y=3%2Ax%2B14(green) that is perpendicular to the given graph and goes through (-3,5)





Now let's convert y=3x%2B14 to standard form


y-3x=14 Subtract 3x from both sides


-3x%2By=14 Rearrange the terms



So the equation that is perpendicular to x%2B3y=6 and goes through (-3,5) is

-3x%2By=14