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

2 Answers
Mar 13, 2017

y=3/2x+9/2

Explanation:

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]}

Mar 13, 2017

3x-2y+9=0.

Explanation:

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.