Question 715519
The x- and y-intercepts of the line with equation
{{{4x+3y-12=0}}} <--> {{{4x+3y=12}}} are easy to find.
For {{{x=0}}} --> {{{4*0+3y=12}}} --> {{{0+3y=12}}} --> {{{3y=12}}} --> {{{y=4}}}
gives you point (0,4) as the y-intercept.
For {{{y=0}}} --> {{{4x+3*0=12}}} --> {{{4x+0=12}}} --> {{{4x=12}}} --> {{{x=3}}}
gives you point (3,0) as the x-intercept.
Those 2 points and point (0,0) , the origin, are the vertices of your triangle:
{{{drawing(300,300,-1,5,-1,5,
grid(1), red(line(-3,8,6,-4))
)}}}
You could say that the length of the base of your triangle is {{{b=3}}} units
and the height is {{{h=4}}} units,
so the area (in square units) is
{{{b*h/2=3*4/2=highlight(6)}}}