Find the equation of the line that contains the point (-5,-3) and is parallel to the line 5x+6y=11. Write the equation in slope-intercept form, if possible.

y = -5x/6 - 33/6

Step-by-step explanation:

The standard form of equation of a line in point slope form is y - y0 = m(x - x0)

m is the slope

(x0, y0) is the point on the line

First we need the slope of the given equation 5x+6y=11.

Rewrite in standard form;

5x+6y=11

6y = -5x + 11

y = -5/6 x + 11/6

Hence the slope is -5/6

Since the unknown line is parallel to this line, they will have the same slope, hence the slope of the required line is -5/6

Substitute m = -5/6 and the point (-5,-3) into the point-slope expression;

y - y0 = m(x - x0)

y - (-3) = -5/6 (x - (-5))

y + 3 = -5/6(x+3)

6(y+3) = -5(x+3)

6y+18 = -5x - 15

6y+5x =-15 -18

6y+5x =-33

Express in slope intercept form

6y = -5x - 33

y = -5x/6 - 33/6

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

step-by-step explanation: