Question 220770
A line perpendicular to x + 3y = 5 passes through (1,-1). What is the equation of the line in standard form.
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Arrange the equation to slope intercept form to find the slope (m1)
x + 3y = 5
subtract x from both sides
3y = -x + 5
Divide both sides by 3
y = {{{-1/3}}}x + {{{5/3}}}
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The slope m1 = {{{-1/3}}}
The relationship of perpendicular lines is given by m1*m2 = -1
we know m1 = {{{-1/3}}}, find m2
{{{-1/3}}}*m2 = -1
Multiply both sides by -3
m2 = -3(-1)
m2 = +3
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Use the point/slope formula to find the equation of the perpendicular line
y - y1 = m(x - x1)
Assign the values as follows: m=3; x1 = 1; y1 = -1
y - (-1) = 3(x - 1)
y + 1 = 3x - 3
y = 3x - 3 - 1
y = 3x - 4
Put in the standard form 
3x - y = 4; is the equation of the perpendicular line
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Did this make sense to you, any questions?