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Let TU be the directed line segment beginning at point T(4,4) and ending at point U(-11,13). Find the point P on the line segment that partitions the line segment into the segments TP and PU at a ratio of 5:1.

Respuesta :

vlatkostojanovic

Answer:

P(-8.5, 11.5)

Step-by-step explanation:

From Exercise we have

T(4,4)  ⇒ x_1=4, y_1=4

U(-11,13)  ⇒  x_2=-11, y_2=13

TP:PU=5:1=a:b ⇒ a=5, b=1

P(x_p, y_p)=?

We have a formula

x_p=x_1+\frac{a}{a+b} · (x_2-x_1)

we get

x_p=4+\frac{5}{6} ·(-11-4)

x_p=4+\frac{5}{6} ·(-15)

x_p=4-75/6

x_p=-8.5

We have a formula

y_p=y_1+\frac{a}{a+b} · (y_2-y_1)

we get

y_p=4+\frac{5}{6} ·(13-4)

y_p=4+\frac{5}{6} ·(9)

y_p=4 + 45/6

y_p=11.5

Therefore, we get P(-8.5, 11.5).

We use the site geogebra.org for drawn the graph.

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