Find an equation of the line that is perpendicular to 10 x + 11 y = 10 and passes through ( − 4 , 8 )

y = (11/10)x + 62/5

Step-by-step explanation:

First find the slope of the given line 10 x + 11 y = 10.  To do this, solve this equation for y and take the slope of the line from the result:

11y = 10 - 10x, or y = (-10/11)x + C, where C is just a constant.  The slope of the given line is (-10/11).

The slope of any line perpendicular to the one above has a slope which is the negative reciprocal of the slope of the given line:

The negative reciprocal of (-10/11) is (11/10).

Starting with the slope-intercept form y = mx + b, and substituting the knowns, we are able to find the value of b:

y = mx + b => 8 = (11/10)(-4) + b, or

                      8 = -44/10 + b, or

                      80 = -44 + 10b

Then 10 b = 124, and so b = 62/5

The desired equation is thus   y = (11/10)x + 62/5