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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.

User Geobits
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6 votes

Final answer:

The equation of the line that is parallel to 2x+3y=3 and passes through the point (3, −4) is y = −(2/3)x − 2. We found this by first converting the given line into slope-intercept form to find the slope, and then using the point-slope form with the given point.

Step-by-step explanation:

To find the equation of the line that is parallel to 2x+3y=3 and passes through the point (3, −4), we first need to determine the slope of the given line. The equation of a line in slope-intercept form is y = mx + b, where m represents the slope. We can rewrite the given equation in slope-intercept form by isolating y:

2x + 3y = 3

3y = −2x + 3

y = −(2/3)x + 1

This shows that the slope (m) of the given line is −(2/3). Since parallel lines have the same slope, the slope of the new line will also be −(2/3). Now we use the point-slope form of a line which is y − y1 = m(x − x1), where (x1,y1) is a point on the line. Plugging in the given point (3, −4) and the slope −(2/3), we get:

y − (−4) = −(2/3)(x − 3)

Distributing the slope on the right side and moving −4 to the other side gives us the final equation:

y = −(2/3)x + 2 + −4

y = −(2/3)x − 2

This is the equation of the line that is parallel to 2x+3y=3 and passes through the point (3, −4).

User Absinthe
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3 votes
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
User Bojan Ilievski
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8.1k points

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