Question 940108
1.

What we know about the line whose equation we are trying to find out:

    it goes through point ({{{15}}}, {{{-9}}})

    it has a slope of {{{m=-(2/5)}}}

the formula for the equation, given point is

{{{y=mx+(y[1]-mx[1])}}} ...plug in given point and slope

{{{y=-(2/5)x+(-9-(-2/5)15) }}}

{{{y=-(2/5)x+(-9-(-2/cross(5))cross(15)3) }}}

{{{y=-(2/5)x+(-9-(-2)3) }}}

{{{y=-(2/5)x+(-9-(-6)) }}}

{{{y=-(2/5)x+(-9+6) }}}

{{{y=-(2/5)x-3 }}}


{{{drawing( 600,600, -20, 20, -20, 20,circle(15,-9,.2),locate(15,-9,p(15,-9)), graph( 600,600, -20, 20, -20, 20,-(2/5)x-3)) }}}


2. What is the equation of the line in point-slope form

{{{x-intercept=-(1/2)}}} => point ({{{-(1/2)}}},{{{0}}})
and 
{{{y-intercept =5}}}=> point ({{{0}}},{{{5}}})


the "point-slope" form:

{{{y – y[1] = m(x – x[1])}}}

{{{y – y[1] = m(x – x[1])}}} if point ({{{-(1/2)}}},{{{0}}})

{{{y – 0 = m(x – (-1/2))}}}

{{{y  = m(x +1/2)}}}..............eq.1


{{{y – y[1] = m(x – x[1])}}} if point ({{{0}}},{{{5}}})=>{{{b=5}}}

{{{y – 5 = m(x – 0)}}}

{{{y -5 = mx }}}

{{{y= mx+5 }}}..............eq.2

since eq.1 and eq.2 have same left sides, right sides must be equal too; so, we have

{{{m(x +1/2)=mx+5}}}

{{{m(x +1/2)-mx=5}}}

{{{m((x +1/2)-x)=5}}}

{{{m(cross(x) +1/2-cross(x))=5}}}

{{{m(1/2)=5}}}

{{{m=5/(1/2)}}}

{{{highlight(m=10)}}}


your equation is:

{{{y=10x+5}}}


{{{drawing( 600,600, -10, 10, -10, 10,
circle(-1/2,0,.15),locate(-1/2,0,p(-1/2,0)),
circle(0,5,.15),locate(0,5,p(0,5)), graph( 600,600, -10, 10, -10, 10,10x+5)) }}}