Question

What are the x- and y- coordinates of point P on the directed line segment from A to B such that P is Two-thirds the length of the line segment from A to B? x = (StartFraction m Over m n EndFraction) (x 2 minus x 1) x 1 y = (StartFraction m Over m n EndFraction) (y 2 minus y 1) y 1 (2, â€"1) (4, â€"3) (â€"1, 2) (3, â€"2).

Answer 1

The x - and y - coordinates of point P are ( - 1 , 2 )

The standard form of the linear equation is,

$y=mx+c$

Where $m$ is the gradient of the line.

Given-

P is 2/3 the length of the line segment from A to B. Suppose $m$ is two, then

$m+n=3$

This gradient of the line can be calculated with the following formula-

$m=\dfrac{y_{2}-y_{1} }{x_{2}- x_{1} }$

Here,  ($x_{1} ,y_{1}$) and ($x_{2} ,y_{2}$) are two arbitrary points.

In the given graph, at point A(9,-8),

$x_{1} =9$

$y_{1} =-8$

At point B (-6,7),

$x_{2} =-6$

$y_{2} =7$

Use the formula has given in the problem which is,

$x=\dfrac{m}{m+n}(x_{2} -x_{1} )+x_{1}$

Put the values,

$x=\dfrac{2}{3}(-6 -9 )+9$

$x=\dfrac{2}{3}(-15 )+9$

$x=2(-5 )+9$

$x=-10+9$

$x=-1$

For y use the formula given in the problem which is

$y=\dfrac{m}{m+n}(y_{2} -y_{1} )+y_{1}$

$y=\dfrac{2}{3}(7 -(-8) )+(-8)$

$y=\dfrac{2}{3}(7+8) )-8$

$y=\dfrac{2}{3}(15) )-8$

$y={2}(5) -8$

$y=2$

Hence, the x - and y - coordinates of point P are ( - 1 , 2 )

For more about the coordinates, follow the link below-

https://brainly.com/question/13140132

Answer 2

Answer: (4, –3)

Step-by-step explanation: