Question

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

Answer 1

y = -2/3x-2

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.

Answer 2

`y = - (2)/(3)x - 2`

EXPLANATION

The given line is

2x+3y=3

Solve for y,

`y = - (2)/(3) x + 1`

The slope of this line is

`m = - (2)/(3)`

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

`m = - (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 = - (2)/(3)(x - 3)`

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

`y = - (2)/(3)x - 2`