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).