Question

On a coordinate plane, a line is drawn from point A to point B. Point A is at (9, negative 8) and point B is at (negative 6, 7).
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, –1)
(4, –3)
(–1, 2)
(3, –2)

Answer 1

Option D: $(3,-2)$ is the coordinates of the point P

Explanation:

The coordinate of point A is $(9,-8)$

The coordinate of point B is $(-6,7)$

The length of the point P is $\frac{2}{3}$ of the line segment from A to B.

The coordinates of the point P can be determined using the formula,

$x=\frac{m}{m+n} (x_2-x_1)+x_1$  and  $y=\frac{m}{m+n} (y_2-y_1)+y_1$

where $m=2,n=3$.

Substituting the values, we have,

$x=\frac{m}{m+n} (x_2-x_1)+x_1$

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

$x=\frac{2}{5} (-15)+9$

$x=-6+9$

$x=3$

Similarly, substituting the values for y, we get,

$y=\frac{m}{m+n} (y_2-y_1)+y_1$

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

$y=\frac{2}{5} (15)-8$

$y=6-8$

$y=-2$

Thus, the coordinates of the point P is $(3,-2)$

Hence, Option D is the correct answer.

Answer 2

Answer:

c

Step-by-step explanation:

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