Question

Find the coordinates of a point that divides a line segment AB in the ratio 2:6.

Answer 1

Explanation:

The first thing we should do is to calculate the distance AB

To do that we must khow the coordinates of the vector AB

• Vector AB(6-(-6)),-7-9)⇒ Vector AB (12, -16)
• So the distance AB =
  `\sqrt{12^(2)+(-16)^(2) }`= 20
• now 20 to the ratio of 2:6
• add 2 and 6 2+6=8
• divide 20 by 8 20/8=2.5
• multiplay 2.5 by 2 and 6 2.5*2=5 and 2.5*6= 15
• so the point (3, -3) can satisfy this ration
• if we calculate the remaining distance we find 5 and 15

Answer 2

Answer: (-3,5)

Explanation:

Percent Ratio =2/2+6 =2/8 =1/4

Rise = −7 − 9 = −16, Run = 6 − (−6) = 6 + 6 = 12

x coordinate of P = x1 + Run(Percent Ratio)

x1 is the x coordinate of the starting point (A) of the line segment

x coordinate of P = −6 + 12(1/4) = −6 + 3 = −3

y coordinate of P = y1 + Rise(Percent Ratio)

y1 is the y coordinate of the starting point (A) of the line segment

y coordinate of P = 9 + (−16)(1/4)) = 9 − 4 = 5

The coordinates of the point that divides line segment AB in the ratio 2:6 are (−3,5).