Question

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.

Answer 1

y = -5x/6 - 33/6

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