write an equation for the line containing the indicated points: (8, -3) and (-8, 3)?

Answer 1

Equation to the line connecting points
(8, --3) and (--8, 3) is
y + 3 = [(3 + 3)/(--8 --8)](x --8)
y + 3 = (--6/16)(x --8)
8y + 24 = --3x + 24
3x + 8y = 0

Answer 2

equation of a line formula y=mx + c
where x and y are co ordinates on the line
m is the slope/gradient of the line
c is where the line cuts the y-axis
to your question first you need to calculate the gradient of the line you need to write its equation using the indicated points.
m=change in y-axis/change in x-axis
m=3-(-3)/-8-8
m=6/-16
m= -3/8
after calculating for the gradient use the formula to find the y-intercept where the line crosses the y-axis.
take one of the indicated points to help you
y=mx+c
taking point (8,-3)
-3=-3/8*8 +c
-3=-3 + c
c= o
so the line crosses at the origin.
so the equation of the line is
y= -3/8x
since c = o.

Answer 3

slope = (3+3) / (-8 -8) = - 6 / 16 = -3 / 8

-3 / 8 = (y +3) / (x - 8)
-3 (x - 8) = 8(y +3)
-3 x + 24 = 8 y + 24
8y - 3x = 0

Answer 4

(y + 3)/(x - 8) = - 6/16
16y + 48 = - 6x + 48
8y = - 3x