Question

What is the equation of a line that is parallel to 4x+6y+5=0 and passes through the points (6,9)

Answer 1

Explanation:

Equation of given line is:

`4x+6y+5=0 \\ \therefore \: 6y = - 4x - 5 \\ \\ \therefore \: y = ( - 4x - 5)/(6) \\ \\ \therefore \: y = ( - 4x)/(6) - (5)/(6) \\ \\ \therefore \: y = - ( 2)/(3)x - (5)/(6)...(1) \\ equating \: eq \: (1) \: with \: y = mx + c, \:we \:\\ find : \\ slope \: of \: given \: line \: (m) = - ( 2)/(3) \\ \because \: required \: line \: is \: \parallel \: to \: given \: line \\ \therefore \: slope \: of \: required \: line \\ = slope \: of \: given \: line \: (m) = - ( 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 = - ( 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`