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