Question 1203163


the straight line {{{y=mx+c }}}passes through the points ({{{0}}}, {{{k}}}) and ({{{h}}},{{{2k}}})




{{{m=(y2-y1)/(x2-x1)}}}

given : ({{{x1}}},{{{y1}}})=({{{0}}},{{{k}}}) and ({{{x2}}},{{{y2}}})=({{{h}}},{{{2k}}}) 

{{{m=(2k-k)/(h-0)}}}

{{{m=k/h}}}


({{{0}}},{{{k}}}) is the {{{y}}}-intercept, so {{{c = k}}}

substitute {{{c}}} in{{{ y=(k/h)x+c }}}

and your equation is:

{{{y=(k/h)x+k }}}