Question 1160656


{{{2y = kx + h}}} passes through the points ({{{-3}}},{{{6}}}) and ( {{{1}}},{{{11}}}) 

find the value of {{{k}}} and of{{{ h}}}


{{{2y = kx + h}}} use  the point ({{{-3}}},{{{6}}})

{{{2*6 = k(-3) + h}}}........solve for {{{h}}}

{{{12 = -3k + h}}}

{{{h=3k+12}}}............eq.1

use the point ( {{{1}}},{{{11}}})

{{{2*11 = k(1) + h}}}.....solve for {{{h}}}

{{{22 = k + h}}}

{{{h=22-k}}}............eq.2


from eq.1 and eq.2 we have:

{{{3k+12=22-k}}}
{{{3k+k=22-12}}}
{{{4k=10}}}
{{{k=10/4}}}
{{{k=5/2}}}

go to {{{h=22-k}}}............eq.2, plug in {{{k}}}

{{{h=22-5/2}}}
{{{h=39/2}}}

then your equation is:


{{{2y = (5/2)x + 39/2}}}

and in slope intercept form is:

{{{y = (5/4)x + 39/4}}}


{{{drawing( 600, 600, -15, 15, -15, 15, 
circle(1,11,.13),circle(-3,6,.13),
locate(1,11,p(1,11)),locate(-3,6,p(-3,6)),
 graph( 600, 600, -15, 15, -15, 15, (5/4)x + 39/4)) }}}