The point is given (6,14) and equation parallel to the equation passes through the point is given Y=4/3X-4.
To determine the equation of line that passes through the point (6,14)
Use the general equation of line.
[tex]y-y_1=m(x-x_1)[/tex]Substitute the values.
[tex]y-14=m(x-6)[/tex]Now the slope of the line will be same as the slope of the equation of line which is parallel .
Because when two lines are parallel the slope of the lines are equal.
[tex]m=\frac{4}{3}[/tex]Then the equation formed is
[tex]\begin{gathered} y-14=\frac{4}{3}(x-6) \\ y=\frac{4}{3}x-8+14 \\ y=\frac{4}{3}x+6 \end{gathered}[/tex]
Hence the equation of line formed is
[tex]y=\frac{4}{3}x+6[/tex]