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Line segment JK in the coordinate plane has endpoints with coordinates \left(-4,11\right)(−4,11) and \left(8,-1\right)(8,−1).Supposed J, P, and K are collinear on segment JK, and JP:JK=\frac{1}{3}JP:JK=1/3 . What are the coordinates of P?

Respuesta :

abidemiokin

Answer:

(-1, 8)

Step-by-step explanation:

Given P to be the point that divides the coordinates J and K in the ratio 1:3

then the coordinate of point P will be xpressed as;

P(X, Y) = [tex](\frac{bx_1+ax_2}{b+a}, \frac{by_1+ay_2}{b+a})[/tex]

Given J = (-4, 11) and K = (8, -1)

[tex]x_1 = -4, y_1 = 11, x_2 = 8 \ and \ y_2 = -1, a = 1, b = 3[/tex]

substituting the given parameters into the line division formula

[tex]P(X, Y) = (\frac{3(-4)+1(8)}{3+1}, \frac{3(11)+(1)(-1)}{3+1})\\\\P(X, Y) = (\frac{-12+8}{4}, \frac{33-1}{4})\\\\P(X, Y) = (\frac{-4}{4}, \frac{32}{4})\\\\P(X, Y) = (-1, 8)[/tex]

Hence the coordinates of P is (-1, 8)

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