The equation of the line in point-slope form is y = -1/3 x + 20/3 and 3y + x = 20
The equation of line perpendicular to another line in point slope form is expressed as;
y-y₁ = -1/m(x-x₁)
Given the equation of a line x -3y - 8 = 0
x - 3y = 8
-3y = -x + 8
y = 1/3x - 8/3
Slope m = 1/3
Substitute the point (-7, 9) and slope 1/3 into the equation
y-9 = -1/3(x-(-7))
y-9 = -1/3(x+7)
3(y-9) = -(x+7)
3y-27 = -x -7
3y = -x + 20
y = -1/3 x + 20/3
Write in general form.
3y = -x + 20
3y + x = 20
Hence the equation of the line in point-slope form is y = -1/3 x + 20/3 and 3y + x = 20
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