A line passes through (2, -1) and (8, 4).Write an equation for the line in point-slope form. Rewrite the equation in slope-intercept form.

Hope this helps

Step-by-step explanation:

1.Find Slope

2 is x1, 8 is x2, (-1) is y1, 4 is y2

y2 - y1

x2 - x1

4-(-1)    = 5

8-2         6

2. Point- slope form:

y - y1 = m (x - x1)

y - (-1) = (5/6) (x - 2)

in the end

y= 5/6x - 8/3

this is the slope intercept form y=mx +/- b

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

i would have to say it is c.