Question 1060847
Use the point slope form,
{{{y-y[0]=m(x-x[0])}}}
{{{y-(-5)=0(x-3)}}}
{{{y+5=0}}}
{{{y=-5}}}
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So the line would have the form,
{{{y=mx+2}}}
So plug in the point to find {{{m}}},
{{{-4=m(3)+2}}}
{{{3m=-6}}}
{{{m=-2}}}
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{{{-4(5t-1)=2(3-10t)-2}}}
{{{-20t+4=6-20t-2}}}
{{{-20t+4=-20t+4}}}
{{{0=0}}}
True, this is an identity.