Question 1173961
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The given function is {{{f(x) = expr(-2/3)x+1/2}}} which is the same as {{{y = expr(-2/3)x+1/2}}} since y = f(x).


This equation is in the form y = mx+b with
m = -2/3 = slope
b = 1/2 = y intercept


We won't use this given y intercept. We'll only use the slope. 


The negative reciprocal of the given slope is 3/2. We flip the fraction to go from -2/3 to -3/2. We also flip the sign from negative to positive.


Side note: The original slope -2/3 and the perpendicular slope 3/2 multiply to -1. This works for any pair of perpendicular lines as long as neither line is vertical.


So far we found the perpendicular slope to be m = 3/2


We want this perpendicular line to go through (x,y) = (2,1)


We will use
m = 3/2
x = 2
y = 1
to plug into y = mx+b. Then we'll solve for b 


So,
y = mx+b
1 = (3/2)*2+b
1 = 3+b
1-3 = b
-2 = b
b = -2


This means y = mx+b updates to {{{y = expr(3/2)x-2}}} to be the equation of the perpendicular line.


If you wish to use point-slope form, then you could do it like this
{{{y - y[1] = m(x - x[1])}}}


{{{y - 1 = (3/2)(x - 2)}}}


{{{y - 1 = (3/2)x - (3/2)*2}}}


{{{y - 1 = (3/2)x - 3}}}


{{{y - 1+1 = (3/2)x - 3+1}}}


{{{y = (3/2)x - 2}}}


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Answer:  {{{y = expr(3/2)x-2}}}


Because 3/2 = 1.5, the decimal form of this equation is {{{y = 1.5x-2}}}
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