Write the equation of the line parallel or perpendicular to the given line that passes through the given point. give your answer in point slope form and slope intercept form.
1. parallel to y=-4/5x + 1 that passes through (-3, -5)
2. perpendicular to y=1/3x + 3 that passes through (4, 1)

Step-by-step explanation:

1. It is convenient to start with point-slope form, then simplify the result to slope-intercept form. The two forms of the equation for a line are ...

  y -k = m(x -h) . . . . . line with slope m through point (h, k)

  y = mx +b . . . . . . . . line with slope m and y-intercept b

The given lines are in slope-intercept form, so we can read the slope directly from the equation.

The slope of the parallel line will be the same as the slope of the given line: -4/5.

  point-slope form: y +5 = (-4/5)(x +3)

  slope-intercept form: y = (-4/5)x -37/5

__

2. The slope of the given line is 1/3. The slope of the perpendicular line is the negative reciprocal of that: -1/(1/3) = -3. Your lines are then ...

  point-slope form: y -1 = -3(x -4)

  slope-intercept form: y = -3x +13

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!