Question 1009436
Those two points could serve as segment for the base; understanding Distance Formula, and area formula of a triangle assumed, the length of that base for those two given points is {{{sqrt(17)}}}.  The altitude of the triangle would be therefore {{{h=(4/17)sqrt(17)}}}.


Consider two line equations.
The base is on {{{y=x/4+7/2}}} and another line to contain the unknown triangle's vertex is some {{{y=x/4+b}}}, and you do not yet know b.


You want the distance between the two lines to be  {{{(4/17)sqrt(17)}}}.  Putting this as an equation,  {{{highlight_green(sqrt((x-x)^2+(x/4+7/2-(x/4+b))^2)=(4/17)sqrt(17))}}}.


Understand that these are two parallel lines, and once you find b, ANY point on the line of the then found & solved b, will contain an acceptable "third" vertex, because any of them will be  the h distance from the chosen base segment.



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(Strategy as as much work as shown, done on paper - not finished beyond how was described here.)


Note that {{{b=(119+- 8sqrt(17))/34}}}.