What is the equation of a line that is parallel to 2x+3y=3 and passes through the point (3, −4) ? Enter your answer in the box.

3y=-2x+3
y=-2/3x+3/3
y=-2/3x+1
compare y=mx+c
m=-2/3
for parallelism m=M
so M=-2/3

now eqn for (3,-4)
is given by y-(-4)/x-3 = -2/3
y+4/x-3=-2/3
3(y+4)=-2(x-3)
3y+12=-2x+6
3y+2x+12-6=0
3y+2x+6=0
or
3y+2x=-6

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erinna erinna · Answer 2 · 2021-10-22T10:42:05+02:00

The equation of the required line is [tex]y=-\dfrac{2}{3}(x)-2[/tex].

Given:

The eqation of paralle line is [tex]2x+3y=3[/tex].

The required line passes through the point [tex](3,-4)[/tex].

To find:

The equation of the required line.

Explanation:

The given equation can be rewritten as:

[tex]3y=3-2x[/tex]

[tex]\dfrac{3y}{3}=\dfrac{3-2x}{3}[/tex]

[tex]y=1-\dfrac{2x}{3}[/tex]

On comparing this equation with [tex]y=mx+b[/tex], we get

[tex]m=-\dfrac{2}{3}[/tex]

Slopes of parallel lines are always equal.

The slope of reqired line is [tex]m=-\dfrac{2}{3}[/tex] and it passing through the point [tex](3,-4)[/tex]. Using the point slope form, the equation of the line is:

[tex]y-y_1=m(x-x_1)[/tex]

[tex]y-(-4)=-\dfrac{2}{3}(x-3)[/tex]

[tex]y+4=-\dfrac{2}{3}(x)-\dfrac{2}{3}(-3)[/tex]

[tex]y=-\dfrac{2}{3}(x)+2-4[/tex]

[tex]y=-\dfrac{2}{3}(x)-2[/tex]

Therefore, the equation of the required line is [tex]y=-\dfrac{2}{3}(x)-2[/tex].

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