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 lo

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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 lo

cated 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.

Mathematics

1 answer:

tester [12.3K] · Answer 1 · 2026-04-09T11:50:43+03:00

The x-axis is nearer to point B. Since the ratio PB:AB = 2/7 is positive, this indicates internal division of the line. We utilize the coordinate formula for a point dividing a line internally. Thus, x = (mx₂ + nx₁)/(m + n) and y = (my₂ + ny₁)/(m + n), where PB/AB = 2/7 translates to n/m, giving n = 2 and m = 7. With point A at (0,0) and point P at (10, -5), we substitute these coordinates to find x and y. The calculations yield coordinates for B at 7⁷/₉ and -3⁸/₉. Therefore, B is -3⁸/₉ units from the x-axis and 7⁷/₉ from the y-axis, confirming that the x-axis is closer to B.