It is a bit tedious to write 6 equations, but it is a straightforward process to substitute the given point values into the form provided.
For segment ab. (x1, y1) = (1, 1); (x2, y2) = (3, 4).
... x = 1 + t(3-1)
... y = 1 + t(4-1)
ab = {x=1+2t, y=1+3t}
For segment bc. (x1, y1) = (3, 4); (x2, y2) = (1, 7).
... x = 3 + t(1-3)
... y = 4 + t(7-4)
bc = {x=3-2t, y=4+3t}
For segment ca. (x1, y1) = (1, 7); (x2, y2) = (1, 1).
... x = 1 + t(1-1)
... y = 7 + t(1-7)
ca = {x=1, y=7-6t}
2 days late, but the answer is no solution. solving a system of equations means finding where they intersect, but by looking at these equations, you know that they never intersect--they're parallel.
they share a slope (2), making them either parallel or "the same line", but the different x-intercepts (9 and -9) mean that they're different lines. they have no solution, or no intersection point, because they're parallel lines.
Answer:
b:100'
a:75
b:50
Step-by-step explanation:
Answer:
D
Step-by-step explanation:
Answer:
2x+6
Step-by-step explanation:
