Step-by-step explanation:
Equation of given line is:
[tex]4x+6y+5=0 \\ \therefore \: 6y = - 4x - 5 \\ \\ \therefore \: y = \frac{ - 4x - 5}{6} \\ \\ \therefore \: y = \frac{ - 4x}{6} - \frac{5}{6} \\ \\ \therefore \: y = - \frac{ 2}{3}x - \frac{5}{6}...(1) \\ equating \: eq \: (1) \: with \: y = mx + c, \:we \:\\ find : \\ slope \: of \: given \: line \: (m) = - \frac{ 2}{3} \\ \because \: required \: line \: is \: \parallel \: to \: given \: line \\ \therefore \: slope \: of \: required \: line \\ = slope \: of \: given \: line \: (m) = - \frac{ 2}{3} \\ \because \: required \: line \: passes \: through \: \\ point \: (6, \: 9) \\ \therefore \: \: equation \: of \: line \: in \: slope \: point \: \\ form \: is \: given \: as: \\ y-y_1=m(x-x_1) \\ \therefore \:y - 9 = - \frac{ 2}{3}(x - 6) \\ \therefore \:3(y - 9 )= - 2(x - 6) \\ \therefore \:3y - 27= - 2x + 12 \\ \therefore \:2x + 3y - 27 - 12 = 0 \\ \purple{ \boxed{\therefore \:2x + 3y - 39 = 0}} \\ is \: the \: equation \: of \: required \: line[/tex]
