Points A, P, and B are collinear on AB, and PB:AB = 2/7. A is located on the origin (0,0), P is located at (10, -5), and B is located at (x, y). Which axis is closer to B, the x-axis or the y-axis? Justify your answer. I don't know how to show the work.

The x-axis is closer to B.

Step-by-step explanation:

Since the ratio PB:AB = 2/7 is positive, the line divides internally. So, we use the equation for the coordinates of a point that internally divide a line.

So, x = (mx₂ + nx₁)/(m + n) and y = (my₂ + ny₁)/(m + n)

where PB/AB = 2/7 = n/m. So, n = 2 and m = 7

Also, Point A = (0,0) = (x₁, y₁) and point P = (10, -5) = (x₂, y₂).

Substituting these values into x and y, we have

x = (mx₂ + nx₁)/(m + n)

x = (7(10) + 2(0)₁)/(7 + 2)

x = (70 + 0)/9

x = 70/9

x = 7⁷/₉

y = (my₂ + ny₁)/(m + n)

y = (7(-5) + 2(0)₁)/(7 + 2)

y = (-35 + 0)/9

y = -35/9

y = -3⁸/₉

So, the x and y coordinates of B are  7⁷/₉ and  -3⁸/₉ respectively. So, point B is -3⁸/₉ units away from the x-axis and 7⁷/₉ units away from the y-axis. So the x-axis is closer to B.

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