Question

parallel and perpendicular lines

Determine an equation of each of the following lines .
*the line perpendicular to y-4=0 and passing through the point (-1,6)

** the line perpendicular to 3x-12y+16=0 and having the same y-intercept as the line 14x-13y-52=0

and explain ....

Answer 1

Put y-4 = o into the form
y = m x + b
y = 0 x + 4
now the slope of the line perpendicular to that is -1/m = -1/0 which is undefined so we will have to use our heads.
y = 4 is a horizontal line through y = 4, x = anything.
x = constant, any constant, is perpendicular to y = 4
So we want the vertical line x = -1
That line, x = -1 , passes through all values of y including 6
---------------------------------------
Now
3x -12 y + 16 = 0
12 y = 3 x + 16
y = (1/4)x + 4/3
what is slope of perpendicular?
m' = -1/(1/4) = -4
so the line we want is
y = -4 x + b
what is b?
well when x = 0, -13 y = 52
y = -52/13
so
-52/13 = -4(0) + b
b = -52/13
so
y = -4x -52/13

Answer 2

Y=-1/4x+8