```Question 399387
1.

*[invoke calculating_slope -1, 2, 5, 2]

answer: {{{y - x - 2 = 0}}} or in slope-intercept form {{{y = x + 2}}}

2.

*[invoke find_line_by_slope_and_point 0, 8, -2]

answer: {{{2x + y - 8 = 0}}} or in slope-intercept form {{{y =-2x+8}}}

3.

{{{y - x = 5}}}  or  {{{- x + y = 5}}}

*[invoke describe_linear_equation -1, 1, 5]

4.

*[invoke find_equation_of_line 1, 1, 0, 2]

answer: {{{x + y - 2 = 0}}}or in slope-intercept form {{{y =-x+2}}}

5.

if slope of the one line is {{{m}}}, the slope of perpendicular line will be {{{m1=-(1/m)}}}

so, if {{{m=1/3}}}, the slope of perpendicular line will be {{{m1=-(1/(1/3))=-3}}}

6.

equation {{{ y = 2x + 5}}} is already in slope-intercept form and you can see that slope is {{{2}}}; so, the slope of a line perpendicular to this line will be {{{-(1/2)}}}

7.

*[invoke find_line_by_slope_and_point -3, 0, 0]

8.

the equation of a line that passes through point (1, -2) and has a slope of 1/3

*[invoke find_line_by_slope_and_point 1, -2, 1/3]

you got:

{{{3x - y - 7 = 0}}}...or in slope-intercept form {{{y=3x-7}}}

{{{x - 3y + 7 = 0}}}...or in slope-intercept form {{{3y=-x-7}}}

or{{{y=-(1/3)x-7/3}}}

{{{x - 3y - 7 = 0 }}}...or in slope-intercept form {{{3y=x-7}}} or{{{y=

(1/3)x-7/3}}}

so the answer is : {{{y=(1/3)x - 7/3}}}

9.

*[invoke find_line_by_slope_and_point 0, 3, 1]

so the answer is : {{{y = x + 3}}}

```