In the rectangular coordinate plane, the coordinates of points A, B, and C are (1, 4), (7, 4 +...

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Mon Jun 15, 2020 4:47 am

Princeton Review

In the rectangular coordinate plane, the coordinates of points A, B, and C are (1, 4), (7, 4 + 6√3) and (7, 4), respectively. What is the absolute value of the difference between AB and BC?

A. 6
B. 4 + √3
C. 6 – 6√3
D. 6√3 – 12
E. 12 – 6√3

OA E

If on a XY-plane, the coordinates of two points are X (x1, y2) and Y(x2, y2), then the distance between the points = XY = √[(x1 – y1)^2 + (x2 – y2)^2]

Thus,

AB = √[1 – 7)^2 + (4 – 4 – 6√3)^2] = √[(–6)^2 + (–6√3)^2] = √(36 + 36*3) = √(36*4) = 6*2 = 12
BC = √{7 – 7)^2 + (4 + 6√3 – 4)^2 = √(6√3)^2 = 6√3

Thus, |AB – BC| = 12 – 6√3

Correct answer: E

Hope this helps!

-Jay
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AAPL wrote: ↑

Mon Jun 15, 2020 4:47 am

Princeton Review

In the rectangular coordinate plane, the coordinates of points A, B, and C are (1, 4), (7, 4 + 6√3) and (7, 4), respectively. What is the absolute value of the difference between AB and BC?

A. 6
B. 4 + √3
C. 6 – 6√3
D. 6√3 – 12
E. 12 – 6√3

OA E

Solution:

Using the distance formula:

AB = √[(7 - 1)^2 + (4 + 6√3 - 4)^2] = √[6^2 + (6√3)^2] = √(36 + 108) = √144 = 12

BC = [(7 - 7)^2 + (4 + 6√3 - 4)^2] = √[0^2 + (6√3)^2] = √[(6√3)^2] = 6√3

Since AB > BC, the absolute value of the difference between AB and BC is 12 - 6√3.

Answer: E