How to determine the equation of the line parallel to 3x - 2y + 4 = 0 and passing through (1,6)?

#color(orange)"Reminder "color(red)(bar(ul(|color(white)(2/2)color(black)("parallel lines have equal slope")color(white)(2/2)|)))#

The equation of a line in #color(blue)"slope-intercept form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=mx+b)color(white)(2/2)|)))#
where m represents the slope and b, the y-intercept.

#"Rearrange "3x-2y+4=0" into this form"#

add 2y to both sides.

#3xcancel(-2y)cancel(+2y)+4=0+2y#

#rArr2y=3x+4#

divide ALL terms on both sides by 2

#(cancel(2) y)/cancel(2)=3/2x+4/2#

#rArry=3/2x+2larr" in form "y=mx+b#

#rArr"slope "=m=3/2#

The equation of a line in #color(blue)"point-slope form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))#
m is slope and # (x_1,y_1)" a point on the line"#

#"For parallel line " m=3/2" and " (x_1,y_1)=(1,6)#

#rArry-6=3/2(x-1)larrcolor(red)" in point-slope form"#

Distributing the bracket and simplifying gives the equation in an alternative form.

#y-6=3/2x-3/2#

#rArry=3/2x-3/2+6#

#rArry=3/2x+9/2larrcolor(red)" in slope-intercept form"#
graph{(y-3/2x-2)(y-3/2x-9/2)=0 [-10, 10, -5, 5]}

Recall that the eqn. of a line parallel to the given line

#l_1:ax+by+c=0# is of the Form #l_2:ax+by+c'=0, c'!=c.#

If we compare the slopes of the lines #l_1 and l_2#, we will find that

the result is quite obvious. If, in addition, #(x_0,y_0) in l_2, then,

#ax_0+by_0+c'=0," giving, "c'=-ax_0-by_0.#

#:. l_2 : ax+by=ax_0+by_0.#

Accordingly, the eqn. of the reqd. line is given by,

#3x-2y=3(1)-2(6) rArr 3x-2y+9=0.#