SOLUTION: Can you please help with this question? Thanks Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3,

Question 87262: Can you please help with this question? Thanks 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) (Show Source): You can put this solution on YOUR website!
First convert the equation 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)

Start with the given equation

Subtract 1x from both sides

Simplify

Divide both sides by 3 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 lets find the perpendicular line

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 (-3,5), 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

Add to 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 (,)

Now convert back to standard form

Solved by pluggable solver: Converting Linear Equations in Standard form to Slope-Intercept Form (and vice versa)

Convert from slope-intercept form (y = mx+b) to standard form (Ax+By = C)

Start with the given equation

Subtract 3x from both sides

Simplify

Multiply both sides by -1 to make the A coefficient positive (note: this step may be optional; it will depend on your teacher and/or textbook)

Distribute and simplify

The original equation (slope-intercept form) is equivalent to (standard form where A > 0)

The equation is in the form where , and

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