Respuesta :

The given line, y-2 = (7/3)(x+5) has a slope of 7/3
It is in the form y-y1 = m(x-x1) where m is the slope

Flip the fraction: 7/3 ---> 3/7
Flip the sign: 3/7 -----> -3/7

The perpendicular slope is -3/7
Let m = -3/7

The given point is (-4,9). So x1 = -4 and y1 = 9

Plug those values into the point slope form equation
y-y1 = m(x-x1)
y-9 = (-3/7)(x-(-4))
y-9 = (-3/7)(x+4)
Answer: Choice C
Hello! The question wants you to write the equation of a line that is perpendicular to the given line. 

The given line 
[tex]y-2 = \frac{7}{3}(x+5)[/tex]

Now we kneed to use the point slope form of a line. Which is as follows.
[tex]y - y_1 = m(x - x_1)[/tex]
m = our slope
The question asks us the equation of the line that is perpendicular to the following line
[tex]y-2 = \frac{7}{3}(x+5)[/tex]
A line is perpendicular when the slope is opposite of the other slope. So this means we flip positive 7/3  to 3/7 and give it a negative sign like so -3/7. Now we have our perpendicular slope, we can now use our point slope form equation.

[tex]y - y_1 = m(x - x_1)[/tex]
Insert our perpendicular slope
[tex]y - y_1 = -\frac{3}{7}(x - x_1)[/tex]
Insert our point (-4,9)
[tex]y - 9 = -\frac{3}{7}(x - (-4))[/tex]
Distribute the - to the -4
[tex]y - 9 = -\frac{3}{7}(x + 4)[/tex] 
The answer is below
[tex]y - 9 = -\frac{3}{7}(x + 4)[/tex] 

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