Question 978968
a) Through the point ({{{2}}},{{{5}}}) with slope {{{m=-3/4}}}


{{{y=mx+b}}}
if slope {{{m=-3/4}}}, we have
{{{y=-(3/4)x+b}}}

now use given point ({{{x}}},{{{y}}})=({{{2}}},{{{5}}}) and find {{{b}}}

{{{5=-(3/4)2+b}}}

{{{5=-(3/2)+b}}}

{{{5+(3/2)=b}}}

{{{b=10/2+3/2}}}

{{{b=13/2}}}

{{{b=6.5}}}

so, your equation is {{{y=-(3/4)x+6.5}}}


 {{{drawing( 600, 600, -10, 10, -10, 10,
circle(2,5,.12), locate(2,5,p(2,5)),
 graph( 600, 600, -10, 10, -10, 10, -(3/4)x+6.5)) }}}




b) through the points ({{{3}}},{{{-1}}}) and ({{{0}}},{{{6}}})

*[invoke change_this_name10094 3, -1, 0, 6]


c) horizontal line through ({{{-2}}},{{{-7}}})

For a horizontal line equation will be {{{y=a}}} where {{{a}}} is a constant.
For the line to pass through the point ({{{-2}}},{{{-7}}}) {{{y}}} should be equal to {{{-7}}}.

Your line should be
{{{y=-7}}}

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-2,-7,.12), locate(-2,-7,p(-2,-7)),
 graph( 600, 600, -10, 10, -10, 10, -7,-7)) }}}


d) line through the point ({{{6}}},{{{5}}}) and parallel to the line {{{2x + 3y = 7}}}

{{{3y =-2x+ 7}}}

{{{y =-(2/3)x+ 7/3}}}


*[invoke equation_parallel_or_perpendicular "parallel", "-2/3", "7/3", 6, 5] 


e) line through the point ({{{4}}},{{{1}}}) and perpendicular to the line {{{x + 5y = 1}}}

{{{5y =-x+ 1}}}

{{{y =-(1/5)x+ 1/5}}}

*[invoke equation_parallel_or_perpendicular "perpendicular", "-1/5", "1/5", 4, 1]