Question 239049
First line is y = .15x + .79
Second line is 0.11x-y= -0.85
Solve for y in the second line to get:
y = .11x + .85
These lines are not parallel or perpendicular because the slopes are not the same or negative reciprocals of each other.


The slope-intercept form of a straight line is y = mx + b where m is the slope.
First line  has slope of .15
Second line has slope of .11


These lines will intersect.
Solve these 2 equations simultaneously to find the intersection.


The equations are:

y = .15x + .79
y = .11x + .85


Mulltiply first eqution by 11 and second equation by 15 to get:


11y = 1.65x + 8.69
15y = 1.65x + 12.75


subtract second equation from first to get:


4y = 4.06


divide both sides by 4 to get:


y = 1.015


substitute in first equation original to get:


1.015 = .15x + .79


solve for x to get:


.15x = 1.015 - .79 = .225 which makes x = .225/.15 = 1.5


y = 1.015 and x = 1.5


substitute in second original equation to get:


1.015 = .11x + .85 becomes 1.015 = .11*1.5 + .85 = .165 + .85 = 1.015 which is true.


The two lines intersect at (x,y) = (1.5,1.015)


graph of these two equations is:


{{{graph(600,600,-10,10,-1,2,.15x + .79,.11x + .85,1.015)}}}


horizontal line was put in at y = 1.015 to help you see the intersection point  better.