Vector vec(AB) has a terminal point (7,-9), an x component of 13 , and a y component of -5 . Find the coordinates of the initial point, A. A=(◻,◻)

Define vector vec(AB): The vector vec(AB) is defined by its terminal point B and its components. The x component of vec(AB) is the difference in the x-coordinates of point B and point A. We are given that the x component is 13, so we can write the equation for the x-coordinate of point A as A_x = B_x - 13.Calculate x-coordinate of point A: Given the terminal point B has an x-coordinate of 7, we can substitute this into the equation to find A_x: A_x = 7 - 13 = -6.Calculate y-coordinate of point A: Similarly, the y component of vec(AB) is the difference in the y-coordinates of point B and point A. We are given that the y component is -5, so we can write the equation for the y-coordinate of point A as A_y = B_y - (-5) or A_y = B_y + 5.Combine x and y coordinates of point A: Given the terminal point B has a y-coordinate of -9, we can substitute this into the equation to find A_y: A_y = -9 + 5 = -4.Combining the x and y coordinates of point A, we get A = (-6, -4).

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Vector
vec(AB) has a terminal point
(7,-9), an
x component of 13 , and a
y component of -5 .
Find the coordinates of the initial point,
A.

A=(◻,◻)

Vector $\overrightarrow{A B}$ has a terminal point $(7,-9)$ , an $x$ component of $13$ , and a $y$ component of $-5$ . $\newline$ Find the coordinates of the initial point, $A$ . $\newline$ $A=(\square, \square)$

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Algebra 2

Write equations of cosine functions using properties

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Q. Vector

\overrightarrow{A B}

has a terminal point

(7,-9)

, an

x

component of

13

, and a

y

component of

-5

\newline

Find the coordinates of the initial point,

A

\newline

A=(\square, \square)

Define vector $\vec{AB}$ : The vector $\vec{AB}$ is defined by its terminal point B and its components. The $x$ component of $\vec{AB}$ is the difference in the $x$ -coordinates of point B and point A. We are given that the $x$ component is $13$ , so we can write the equation for the $x$ -coordinate of point A as $A_x = B_x - 13$ .
Calculate x-coordinate of point A: Given the terminal point B has an x-coordinate of $7$ , we can substitute this into the equation to find $A_x$ : $A_x = 7 - 13 = -6$ .
Calculate y-coordinate of point A: Similarly, the y component of $\vec{AB}$ is the difference in the y-coordinates of point B and point A. We are given that the y component is $-5$ , so we can write the equation for the y-coordinate of point A as $A_y = B_y - (-5)$ or $A_y = B_y + 5$ .
Combine $x$ and $y$ coordinates of point $A$ : Given the terminal point $B$ has a $y$ -coordinate of $-9$ , we can substitute this into the equation to find $A_y$ : $A_y = -9 + 5 = -4$ .
Combine $x$ and $y$ coordinates of point $A$ : Given the terminal point $B$ has a $y$ -coordinate of $-9$ , we can substitute this into the equation to find $A_y$ : $A_y = -9 + 5 = -4$ . Combining the $x$ and $y$ coordinates of point $A$ , we get $y$ $1$ .