After 16 years in a savings account, allison has $7,280 in interest. how much was the initial amount if the account paid 7% simple interest?

(–1, 1) and (–6, –1)

(1, 0) and (6, 2)

(3, 0) and (8, 2)

Step-by-step explanation:

Slope of line having points (x1, y1) and (x2,y2) is given by (y2-y1)/(x2-x1)

Two lines with slope m1 and m2 are parallel if there slopes are equal i.e m1 = m2.

Slope of line  that contains (3, 4) and (–2, 2)

m = (2-4)/(-2-3) = -2/-5 = 2/5

To find if other pair of lines lie on line parallel to  line  that contains (3, 4) and (–2, 2), we need to find slope for other pair of points , if those pairs have slopes as 2/5 then its clear that they lie on line parallel to  line  that contains (3, 4) and (–2, 2) else not.

Lets find slope of all the pairs

(–2, –5) and (–7, –3)

m1 = -3 -(-5)/-7-(-2) = -3+5/-7+2

m1 = 2/-5 = -2/5

as  -2/5 is not equal to 2/5, the mentioned points do not  lie on line parallel to  line  that contains (3, 4) and (–2, 2).

(–1, 1) and (–6, –1)

m2 = -1 -1)/-6-(-1) = -2/-6+1

m2 = -2/-5 = 2/5

as  2/5 is  equal to 2/5, the mentioned points do  lie on line parallel to  line  that contains (3, 4) and (–2, 2).

(0, 0) and (2, 5)

m3 = 5 -0)/2-0

m3 = 5/2

as   5/2 is not equal to 2/5, the mentioned points do not lie on line parallel to  line  that contains (3, 4) and (–2, 2).

(1, 0) and (6, 2)

m4 = 2 -0)/6-1

m4 = 2/5

as  m4

As m4 is  equal to 2/5, the mentioned points do  lie on line parallel to  line  that contains (3, 4) and (–2, 2).

(3, 0) and (8, 2)

m5 = 2 -0)/8-3

m5 = 2/5

as  m5 is  equal to 2/5, the mentioned points do  lie on line parallel to  line  that contains (3, 4) and (–2, 2).

Thus, pair of points which lie on  lie on line parallel to  line  that contains (3, 4) and (–2, 2) are

(–1, 1) and (–6, –1)

(1, 0) and (6, 2)

(3, 0) and (8, 2)