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defon
3 years ago
14

Write an equation that is parallel to 3x-2y=14 and passes through point (-6,-11)

Mathematics
1 answer:
Allisa [31]3 years ago
5 0

y = (\frac{3}{2})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 = (\frac{3}{2})x - 7

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

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 \frac{3}{2}.

Step 3 :

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

(y-y_{1}) = m((x-x_{1})

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

(y-(-11)) = \frac{3}{2} ( x - (-6))

y + 11 = \frac{3}{2} ( x + 6)

y = (\frac{3}{2})x + 9 - 11

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

Step 4 :

Answer  :

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

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