Question

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.

Answer 1

Answer:

$P=B=(6,-10)$

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:

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

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

P=B=(6,-10)

Answer 2

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:

$P=\sqrt{(x_{2}-x_{1}) ^{2} + (y_{2}-y_{1}) ^{2}}$

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

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

P ≅ 14.32