Question 260757
First, we must solve for y to get the equation in slope intercept form.



{{{3x + 2y = -1}}} Start with the given equation



{{{2y = -1-3x}}} Subtract 3x from both sides.



{{{2y = -3x-1}}} Rearrange the terms.



{{{y = (-3x-1)/2}}} Divide both sides by 2 to isolate y.



{{{y = (-3/2)x-1/2}}} Break up the fraction and simplify.



So the equation {{{y = (-3/2)x-1/2}}} is now in slope intercept form {{{y=mx+b}}} where the slope is {{{m=-3/2}}} and the y intercept is {{{b=-1/2}}}


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Recall that perpendicular lines have slopes that are negative reciprocals of one another. So because the slope of the given line is {{{-3/2}}}, the perpendicular slope is {{{2/3}}} (just flip the fraction and change the sign).



So we now know that the slope of the perpendicular line is {{{m=2/3}}} and it goes through the point (4,5).



Now let's use the point-slope formula to find the perpendicular equation.



{{{y-y[1]=m(x-x[1])}}} Start with the point-slope formula



{{{y-y[1]=(2/3)(x-x[1])}}} Plug in {{{m=2/3}}}



{{{y-5=(2/3)(x-4)}}} Plug in {{{x[1]=4}}} and {{{y[1]=5}}}. This is given from (4,5). Our goal now is to solve for y.



{{{y-5=(2/3)x-(2/3)4}}} Distribute



{{{y-5=(2/3)x-8/3}}} Multiply



{{{y=(2/3)x-8/3+5}}} Add 5 to both sides.



{{{y=(2/3)x+7/3}}} Combine like terms.


So the equation of the perpendicular line is {{{y=(2/3)x+7/3}}} (make sure that there isn't a typo). Notice that if we graph the two lines, we get


{{{drawing(500,500,-10,10,-10,10,
grid(1),
graph(500,500,-10,10,-10,10,(-1-3x)/2,(2/3)x+7/3),
circle(4,5,0.05),circle(4,5,0.08),circle(4,5,0.10),circle(4,5,0.12)
)}}}


Graph of {{{y=(-3/2)x-1/2}}} (red) and {{{y=(2/3)x+7/3}}} (green) going through the point (4,5). Notice how the red line is perpendicular to the green line.