Let AB be the line segment beginning at point A(0, 3) and ending at point B(6, -10). Find the point P on the line segment that is of the distance from A to B.

Question

contestada

Respuesta :

ukshedrack ukshedrack · Answer 1 · 2020-04-03T02:51:54+02:00

Answer:

Point P = 14.32

Step-by-step explanation:

Point P represents the distance from point A to point B.

The formula is given as:

[tex]P=\sqrt{(x_{2}-x_{1}) ^{2} + (y_{2}-y_{1}) ^{2}}[/tex]

x1 = 0, y1 = 3, x2 = 6, and y2 = -10

[tex]P=\sqrt{(6-0) ^{2} + (-10-3) ^{2}}\sqrt{x} \\\\P = \sqrt{6^{2} + (-13)^{2} }= \sqrt{36+169} \\\\P = \sqrt{205}\\ \\P = 14.3178[/tex]

P ≅ 14.32

Answer Link

profantoniofonte profantoniofonte · Answer 2 · 2020-04-03T04:00:18+02:00

Answer:

[tex]P=B=(6,-10)[/tex]

Step-by-step explanation:

1) Firstly let's place the points in the Cartesian Plane, A is the starting point.

According to the coordinates given:

(Check it out)

2) The distance from A to B, is calculated by:

[tex]d_{AB}=\sqrt{(-10-3)^{2}+(6-0)^2}\cong 14.32\:u[/tex]

The point P on this line segment AB that is of the distance of 14.32 units is B.

P=B=(6,-10)

Answer Link