What equation would represent the same linear relationship for the points (40,80) (80,115) (120,140) (160,165)

You can use the formula y1-y2=m(x1-x2)

to find m, you just do (y2-y1)/(x2-x1) so

(115-80)/(80-40)= 35/40= 7/8

then plug it into the point slope equation:

y-115=7/8(x-80)

y-115=7/8x-70
y= 7/8 x + 45

Assuming they are linear
they should have the same slope and y intercept
y-y1=m(x-x1)
m=slope
(x1,y1) is a point on the line

so pick any 2 points
(40,80) and (80,115)
for 2 points (x1,y1) and (x2,y2) the slope is (y2-y1)/(x2-x1)
for (40,80) and (80,115), the slope is (115-80)/(80-40)=35/40=7/8

y-y1=7/8(x-x1)
if we pick (40,80)
(x,y)
x1=40
y1=80
so
y-80=7/8(x-40)
if we expand
y-80=7/8x-35
add 80
y=7/8x+45
captain, we have a problem
the equation for the last 3 points ((80,115) (120,140) (160,165) ) is y=5/8x+65
while the equation for the first 2 points ((40,80) (80,115)) is y=7/8x+45

Arhombus is a quadrilateral.