Respuesta :

y = ([tex]\frac{3}{2}[/tex])x -2 is the equation of the required line

Step-by-step explanation:

Step 1 :

Equation of the given line is 3x-2y=14

Re writing this in the form y = mx + c , we have

-2y = -3x +14

 y = ([tex]\frac{3}{2}[/tex])x - 7

The co efficient of x , m is the slope of the line. So for the given line the slope is  [tex]\frac{3}{2}[/tex]

Step 2 :

We have to find  equation of a line which is parallel to this line. All parallel lines will have the same slope. Hence the required line has a slope of [tex]\frac{3}{2}[/tex].

Step 3 :

Equation of line with slope m and passing through a point ([tex]x_{1} ,y_{1}[/tex]) is

(y-[tex]y_{1}[/tex]) = m((x-[tex]x_{1}[/tex])

So equation of line passing through (-6,-11) and with a slope of  [tex]\frac{3}{2}[/tex] is

(y-(-11)) = [tex]\frac{3}{2}[/tex] ( x - (-6))

y + 11 = [tex]\frac{3}{2}[/tex] ( x + 6)

y = ([tex]\frac{3}{2}[/tex])x + 9 - 11

y = ([tex]\frac{3}{2}[/tex])x -2

Step 4 :

Answer  :

y = ([tex]\frac{3}{2}[/tex])x -2 is the required equation