Question 764907
Line AB is y=-(7/2)x+3, and the line perpendicular to it at point (0,3) is y=(2/7)x+3.  The point (0,3) here was picked arbitrarily for convenience.  


Using segment AB as a base of the triangle, distance formula, steps here omitted,... the base AB is {{{sqrt(53)}}}.  

Triangle with 8 unit^2 will have an altitude value h, so that:
{{{(1/2)(sqrt(53))*h=8}}}
{{{highlight(h=16/sqrt(53))}}}


What point is h units away from (0,3) and is on line y=(2/7)x+3?
That refers to another distance and the use of distance formula again.
We want those two,  variable points of (x, (2/7)x+3).
Applying distance formula,
{{{h=16/sqrt(53)=sqrt((x-0)^2+((2/7)x+3-3)^2)}}}
which will simplify to:
.
.
{{{highlight(x=-112/53)}}} or {{{highlight(x=112/53)}}}
From those two values, you can use {{{y=(2/7)x+3}}} to determine the corresponding values for y.


Can you finish the rest of the way to the needed line {{{y=-(7/2)x+g}}} for some y-intercept g ?



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Just to show final answer without the remaining steps,
lines wanted are:
{{{highlight(y=-(7/2)x+519/53)}}} and {{{highlight(y=-(7/2)x+11)}}}
and know that point C is really still a variable point.