Question 988589
{{{y=3x-1}}} , and {{{x+3y-6=0}}} are the equations of two sides of the square.
The coordinates of the vertex at the intersection of those two sides can be found by solving the system
{{{system(y=3x-1,x+3y-6=0)}}}--->{{{system(y=3x-1,x+3(3x-1)-6=0)}}}--->{{{system(y=3x-1,x+9x-3-6=0)}}}--->{{{system(y=3x-1,10x-9=0)}}}--->{{{system(y=3x-1,x=9/10)}}}--->{{{system(y=3(9/10)-1,x=9/10)}}}--->{{{system(y=27/10)-10/10,x=9/10)}}}--->{{{highlight(system(y=17/10,x=9/10))}}} .
So, point (9/10,17/10), or (0.9,1.7) is one vertex,
the one at the intersection of the two given sides.
Point (0,-1) , with {{{system(x=0,y=-1)}}} is the other end/vertex of that side with equation {{{y=3x-1}}} .
So far we have this:
{{{drawing(300,300,-1,4,-2,3,
grid(0),arrow(-2,-7,2,5),
arrow(-6,4,6,0),red(line(0,-1,0.9,1.7)),
red(circle(0,-1,0.05)),red(circle(0.9,1.7,0.05)),
locate(0.1,-1.1,red(A)),locate(1,1.9,red(B))
)}}}
What we do not know is to what side of segment {{{red(AB)}}} to draw the square.
There are really two right answers. Which one will be the right one according to your textbook? Or did the authors realize that there were two possible answers?
We can draw the square to the right, like this:
{{{drawing(300,300,-1,4,-2,3,
grid(0),
red(line(0,-1,0.9,1.7)),red(line(3.6,0.8,2.7,-1.9)),
red(line(0,-1,2.7,-1.9)),red(line(3.6,0.8,0.9,1.7)),
red(circle(0,-1,0.05)),red(circle(0.9,1.7,0.05)),
red(circle(3.6,0.8,0.05)),red(circle(2.7,-1.9,0.05))
)}}} , or we can draw the square to the left, as Alan suggested.
Either way, if we encase that square inside a larger square with sides parallel to the x- and y-axes,
we form four congruent right triangles with legs measuring
{{{x[B]-x[A]=0.9-0=0.9}}} and {{{y[B]-y[A]=1.7-(-1)=2.7}}} , like this {{{drawing(300,300,-1,4,-3,2,
grid(0),
red(line(0,-1,0.9,1.7)),red(line(3.6,0.8,2.7,-1.9)),
red(line(0,-1,2.7,-1.9)),red(line(3.6,0.8,0.9,1.7)),
red(circle(0,-1,0.05)),red(circle(0.9,1.7,0.05)),
red(circle(3.6,0.8,0.05)),red(circle(2.7,-1.9,0.05)),
line(0,-1.9,0,1.7),line(3.6,1.7,0,1.7),
line(3.6,-1.9,3.6,1.7),line(3.6,-1.9,0,-1.9),
locate(0.05,-1.3,0.9),locate(3.17,1.3,0.9),
locate(2.5,1.7,2.7),locate(1.2,-1.65,2.7),
rectangle(0,-1.9,0.2,-1.7),rectangle(3.6,1.7,3.4,1.5),
rectangle(0,1.7,0.2,1.5),rectangle(3.6,-1.9,3.4,-1.7),
locate(0.05,-1,red(A)),locate(1,1.95,red(B)),
locate(2.65,-1.96,red(F)),locate(3.67,0.9,red(E))
)}}}
So, {{{x[E]=x[B]+2.7=0.9+2.7=3.6}}} , {{{y[E]=y[B]-0.9=1.7-0.9=0.8}}} ,
{{{x[F]=x[A]+2.7=0+2.7=2.7}}} , and {{{y[F]=y[A]-0.9=-1-0.9=-1.9}}} .