hi mariah
i am slightly confused with the word encircled (which imply circle), to me we must find the area restricted with lines
it looks like right triangle, so if we find the legs of triangle we can find its area
it is easy to construct lines x=2, y=3, and y=x-6, and then to find coordinates of the vertex
x=2, and y=x-6 insert x=2 to the y=x-6,y=2-6=-4, so the first leg will equal=3+4=7
y=3, and y=x-6, here insert y=3 into the y=x-6, 3=x-6, x=9 thte second leg will equal 9-2=7 also 7
and now the area=1/2*7*7=49/2

but if we need to find the area if the circle inscribed into the given triangle we must first find radius, r= (leg1+leg2-hypotenuse)/2=(7+7-2*7^1/2)/2=7-7^1/2
S=pi*(7-7^1/2)^2

hi Clock60 you are correct. the area covered should be 49/2 (i found the area using the same way). I an sure encircled would mean enclosed because there cannot be any circle in this question. i mean yes we can find the incircle however then the question would change.

Guys i have a completely different view, The figure formed is actually a square...and IMO the answer is to your surprise 49...moreover the area question is talking about is i think the area enclosed between the lines...as we get 4 lines after solving the inequalities

Explanation:-

As u already found the values...

X max = 9 and X min = 2 (X=9 and X=2 are 2 lines)
Y max = 3 and Y min = -4 (Y=3 and Y=(-4) are 2 lines)

so we get 4 lines

if u draw on X-Y plane, u get a square (shaded region enclosed b/w them) with

side || to X-axis of length = 9-2 = 7
side || to Y-axis of length = 3-(-4) = 7

Hence the enclosed area = 7*7 = 49

correct me if m wrong???

Graph the three lines y=3 x=2 and y=x-6

the first is parallel to the x axis at y = 3, the second is parallel to the y axis at x = 2 and the third has a slope of 1 and a y intercept of -6. To see the relevant figure, shade everything above the line y = 3, everything to the left of the line x = 2 and everything below the line y=x-6. The unshaded area is then "encircled" by those three lines and is, indeed, a triangle.