What are the 6 point that divide the line between (-5,4) and (13,22)?

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What are the 6 point that divide the line between (-5,4) and (13,22)?

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Answer 1 · 2018-04-29T03:57:36

see explanation.

Explanation:

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If the line $P S$ $PS$ is divided into 6 congruent segments, there should be 5 points in between $P S$ $PS$ , namely, $P_{1}, P_{2}, P_{3}, P_{4} and P_{5}$ $P_1,P_2,P_3,P_4 and P_5$ , as shown in the figure.
As $P P_{1}, P_{1} P_{2}, P_{2} P_{3}, P_{3} P_{4}, P_{4} P_{5}, and P_{5} S$ $PP_1, P_1P_2, P_2P_3,P_3P_4,P_4P_5, and P_5S$ are equidistant,
$\Rightarrow x = \frac{13 - (- 5)}{6} = 3$ $=> x=(13-(-5))/6=3$
similarly, $y = \frac{22 - 4}{6} = 3$ $y=(22-4)/6=3$
$\Rightarrow P_{1} = (x_{1}, y_{1}) = (- 5 + x, 4 + y) = (- 5 + 3, 4 + 3) = (- 2, 7)$ $=> P_1=(x_1,y_1)=(-5+x,4+y)=(-5+3,4+3)=(-2,7)$
similarly, $P_{2} = (x_{2}, y_{2}) = (x_{1} + x, y_{1} + y) = (- 2 + 3, 7 + 3) = (1, 10)$ $P_2=(x_2,y_2)=(x_1+x,y_1+y)=(-2+3,7+3)=(1,10)$
$P_{3} = (x_{3}, y_{3}) = (x_{2} + x, y_{2} + y) = (1 + 3, 10 + 3) = (4, 13)$ $P_3=(x_3,y_3)=(x_2+x,y_2+y)=(1+3,10+3)=(4,13)$
$P_{4} = (x_{4}, y_{4}) = (x_{3} + x, y_{3} + y) = (4 + 3, 13 + 3) = (7, 16)$ $P_4=(x_4,y_4)=(x_3+x,y_3+y)=(4+3,13+3)=(7,16)$
$P_{5} = (x_{5}, y_{5}) = (x_{4} + x, y_{4} + y) = (7 + 3, 16 + 3) = (10, 19)$ $P_5=(x_5,y_5)=(x_4+x,y_4+y)=(7+3,16+3)=(10,19)$

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What are the 6 point that divide the line between (-5,4) and (13,22)?

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