Question 1122738
1.Find an equation of the line that passes through the origin and has a slope of {{{- 5}}}

an equation of the line is {{{y=mx+b}}}, where {{{m}}} is a slope and {{{b}}} is y-intercept

you are given: 

the line that passes through the origin, which means the line that passes through the point ({{{0}}},{{{0}}})

the line that passes through the origin, which means {{{m=-5}}}

go to {{{y=mx+b}}}, substitute what is given

{{{0=-5*0+b}}}...solve for {{{b}}}

{{{b=0}}}

so, your line is: {{{y= -5x}}}

{{{drawing( 600, 600, -10, 10,-10, 10,
circle(0,0,0.12), locate(0.1,0.1,o(0,0)),
 graph( 600, 600, -10, 10, -10, 10, -5x)) }}}


2. Write an equation in standard form of the line satisfying the following condition.
The line goes through ​(3​,2​), and is horizontal.

if given that the line is horizontal, we know that horizontal line has a slope equal ti {{{zero}}};  it must be {{{y=b}}}

in this case {{{b}}} is y coordinate of given point; {{{b=2}}}

so, your equation is: {{{y=2}}}


{{{drawing( 600, 600, -10, 10,-10, 10,
circle(3,2,0.12), locate(3,21,p(3,2)),
 graph( 600, 600, -10, 10, -10, 10, 2)) }}}


3. Write an equation of the line containing ​(2​,-1​) and parallel to the graph of {{{7x-5y=4}}}.

given:

 ​(2​,-1​)

{{{7x-5y=4}}}=>{{{7x-4=5y}}}=>{{{y=(7/5)x-4/5}}}


 *[invoke equation_parallel_or_perpendicular "parallel", "7/5", "4/5", 2, -1]