Point c is located at (1, 2) and point d is located at (−4, −2). find the x value for the point that is  1\4  the distance from point c to point d

The easiest way to do this is to realize this is a weighted average of the two points.  (A weighted average is a linear combination where the coefficients add to one.)

(x,y) = (1-t)C + tD

where t is a real parameter. When t=0 we're at point C, when t=1 we're at point D. We're interested in t=1/4,

(x,y)= (3/4) C + (1/4) D

We're only asked for the x coordinate:

x = (3/4) (1) + (1/4) (-4) = 3/4 - 1 = - 1/4

1/3 + 1/5 = 8/15.

step-by-step explanation:

hope it

graph b

step-by-step explanation:

one solution- graph a (clear intercepting point.)

no solution- graph b (they're parallel, there will never be a point of interception.)

infinite solutions- graph c (all points intercept.)