Hello,
Vertices are on a line parallele at ox (y=-3)
The hyperbola is horizontal.
Equation is (x-h)²/a²- (y-k)²/b²=1
Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)
Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6
c²=a²+b²==>b²=36-4²=20
Equation is:
SA = 2lw + 2lh + 2hw
SA = 2(15)(2) + 2(15)(6) + 2(6)(2)
SA = 60 + 180 + 24
SA = 264
Notice that for this question, they are providing you with:
1) The slope of the line (4)
2) a point on the plane through which the line goes (6, -2)
So you can directly use for this the "point-slope" form of a line which is as follows:

This includes m as the slope, and the pair (x0, y0) as the pair representing the point the line goes through.
So, let's use the info they gave you to complete this:

which we can work out a little more:

And even a little more to finish the equation of the line in its standard slope-intercept form:

So the equation of the line is: