Question

what is the equation of a line that is paralle to 2x+3y=3 and passes through the point (3,-4)​

Answer 1

ANSWER

$y = - \frac{2}{3}x - 2$

EXPLANATION

The given line is

2x+3y=3

Solve for y,

$y = - \frac{2}{3} x + 1$

The slope of this line is

$m = - \frac{2}{3}$

The line that is parallel to this line also has slope,

$m = - \frac{2}{3}$

Since the line passes through (3,-4),

We can use the slope intercept formula,

$y-y_1=m(x-x_1)$

We substitute the slope and the point to obtain,

$y + 4 = - \frac{2}{3}(x - 3)$

$y = - \frac{2}{3} x + 2 - 4$

$y = - \frac{2}{3}x - 2$

Answer 2

Answer:

y = -2/3x-2

Step-by-step explanation:

We have to find the  equation of a line that is paralle to 2x+3y=3 and passes through the point (3,-4)​.

The given equation is:

2x+3y=3

Solving it for y we get,

y = -2/3x+1

The slope of this line is :

m = -2/3

This line is parallel  to the required line so the parallel lines have equal slopes.

The slope of this required line is:

m -2/3

Line passes through points (3,-4)

The standard  slope-intercept form of equation is:

(y-y₁)= m(x-x₁)

So, the equation of the required line is :

y+ 4 = -2/3(x-3)

y = -2/3x+2-4

y = -2/3x-2 is the equation of line.