Question 1016826
Pick any two points to serve as a base for finding the area.


Points  (1,-3) and (3,2):
{{{sqrt((1-3)^2+(-3-2)^2)=sqrt(4+36)=sqrt(40)=sqrt(4*10)=highlight_green(2sqrt(10))}}}.


The line containing those points is...
{{{m=(2-(-3))/(3-1)=5/3}}}
{{{y-2=(5/3)(x-3)}}}
{{{y-2=(5/3)x-5}}}
{{{highlight_green(y=(5/3)x-3)}}}


Use that line, and the not-yet-used vertex point of  (-2,4).
What is equation of the line perpendicular to the chosen base and containing point (-2,4)?
{{{y-4=-(3/5)(x-(-2))}}}
{{{y-4=-(3/5)(x+2)}}}
{{{y-4=-(3/5)x-6/5}}}
{{{y=-(3/5)x-6/5+4}}}
{{{y=-(3/5)x-6/5+20/5}}}
{{{highlight_green(y=-(3/5)x+14/5)}}}


What is the "height" of the triangle, using this point (-2,4) as one endpoint?
Find the intersection point of  {{{y=-(3/5)x+14/5}}}  and  {{{y=(5/3)x-3}}}.
Skipping the steps here, that intersection point is  ( 87/34, 43/34 ).
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The height you want is {{{h=sqrt((87/34-(-2))^2+(43/34-4)^2)}}}.
Simplify this as much as possible.


Your goal is to use area formula  {{{(1/2)(2*sqrt(10))*h}}}.
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To show that the triangle is isosceles, just use Distance Formula between the three segments.