SOLUTION: Find the equation, in standard form, with all interger coeffecients, of the line perpendicular to 3x - 6y = 9 and passing through (-2,-1)

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Question 114588: Find the equation, in standard form, with all interger coeffecients, of the line perpendicular to 3x - 6y = 9 and passing through (-2,-1)
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
First convert 3x - 6y = 9 to 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)


Start with the given equation


Subtract 3x from both sides


Simplify


Divide both sides by -6 to isolate y


Break up the fraction on the right hand side


Reduce and simplify


The original equation (standard form) is equivalent to (slope-intercept form)


The equation is in the form where is the slope and is the y intercept.





Now let's find the equation of the line through points (-2,-1) that is perpendicular to
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 , you can find the perpendicular slope by this formula:

where is the perpendicular slope


So plug in the given slope to find the perpendicular slope



When you divide fractions, you multiply the first fraction (which is really ) by the reciprocal of the second



Multiply the fractions.


So the perpendicular slope is



So now we know the slope of the unknown line is (its the negative reciprocal of from the line ). Also since the unknown line goes through (-2,-1), we can find the equation by plugging in this info into the point-slope formula

Point-Slope Formula:

where m is the slope and (,) is the given point



Plug in , , and



Distribute



Multiply



Subtract from both sides to isolate y

Combine like terms

So the equation of the line that is perpendicular to and goes through (,) is


So here are the graphs of the equations and




graph of the given equation (red) and graph of the line (green) that is perpendicular to the given graph and goes through (,)
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