Respuesta :

arthurpdc
Initially, we'll calculate the slope m of [tex]2x+3y=6[/tex]. Since it's parallel to the line that we want to find, the slope of both are equals. Then:

[tex]2x+3y=6\\\\ 3y=-2x+6\\\\ y=-\frac{2}{3}x+\frac{6}{3}\\\\ y=-\frac{2}{3}x+2\Longrightarrow m=-\dfrac{2}{3}[/tex]

Now, we'll use the equation [tex]y-y_0=m(x-x_0)[/tex] (where [tex]m[/tex] is the slope and [tex](x_0,y_0)[/tex] is a point that the line contains) to find the line:

[tex]y-y_0=m(x-x_0)\\\\ y-(3)=-\dfrac{2}{3}(x-(-2))\\\\ y-3=-\dfrac{2}{3}(x+2)\\\\ 3y-9=-2(x+2)\\\\ 3y-9=-2x-4\\\\ \boxed{r:~~3y+2x=5}[/tex]

3y+2x =5 is the equation of the line.

Initially, we'll calculate the slope m of 2x+3y=6. Since it's parallel to the line that we want to find, the slope of both is equal. Then:

2x+3y =6

3y= -2x+6

y= -2x/3 +6/3

y= -2x/3 +2

m = -2/3

Now, we'll use the equation y-y0 =m(x-x0)  (where  is the slope and  is a point that the line contains) to find the line:

y-y0= m(x-x0)

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

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

3y-9 = -2(x+2)

3y-9 = -2x-4

3y+2x =5

What is the equation of the road example?

The equation, y = MX + b, is in slope-intercept form for the equation of a line. Whilst an equation is in this form, the slope of the road is given via m and the y-intercept is positioned at b. for example, a line with the equation y = 2x + four has a slope of two and a y-intercept of four.

The general equation of a straight line is y = MX + c, wherein m is the gradient, and y = c is the value where the line cuts the y-axis. This wide variety of c is called the intercept on the y-axis. The equation of a straight line with gradient m and intercept c on the y-axis is y = MX + c.

Learn more about the equation of the line here: https://brainly.com/question/13763238

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