Answer: The equation of the line is [tex]3x-y=14.[/tex]
Step-by-step explanation: We are given to find the equation of the line that is perpendicular to 3x + 9y = 7 and passes through the point (6, 4).
The slope-intercept form of the given equation is
[tex]3x+9y=7\\\\\Rightarrow 9y=-3x+7\\\\\Rightarrow y=-\dfrac{1}{3}x+\dfrac{7}{9}.[/tex]
So, slope will be given by
[tex]m=-\dfrac{1}{3}.[/tex]
If 'p' represents the slope of the perpendicular line, then we must have
[tex]m\times p=-1\\\\\Rightarrow -\dfrac{1}{3}\times p=-1\\\\\Rightarrow p=3.[/tex]
Therefore, the equation of the line with slope p = 3 and passing thjrogh the point (6, 4) is given by
[tex]y-4=p(x-6)\\\\\Rightarrow y-4=3(x-6)\\\\\Rightarrow y-4=3x-18\\\\\Rightarrow 3x-y=14.[/tex]
Thus, the required equation of the line is [tex]3x-y=14.[/tex]
