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One of the endpoints of a line segment is located at
(9,8). The midpoint of the line segment is located at
(-3,-16). What are the coordinates of the other endpoint of the line segment?
A
(6,-8)
B.
(3,-4)
C.
(-6,-32)
D.
(-15,-40)

$6$ . One of the endpoints of a line segment is located at $(9,8)$ . The midpoint of the line segment is located at $(-3,-16)$ . What are the coordinates of the other endpoint of the line segment? $\newline$ A $(6,-8)$ $\newline$ B. $(3,-4)$ $\newline$ C. $(-6,-32)$ $\newline$ D. $(-15,-40)$

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Write a quadratic function from its vertex and another point

Full solution

Q.

6

. One of the endpoints of a line segment is located at

(9,8)

. The midpoint of the line segment is located at

(-3,-16)

. What are the coordinates of the other endpoint of the line segment?

\newline

A

(6,-8)

\newline

B.

(3,-4)

\newline

C.

(-6,-32)

\newline

D.

(-15,-40)

Midpoint formula: Midpoint formula: $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$ $\newline$ Given midpoint $(-3, -16)$ and one endpoint $(9, 8)$ .
Given midpoint and endpoint: Let's call the other endpoint $(x_2, y_2)$ .
$(-3, -16) = \left(\frac{9 + x_2}{2}, \frac{8 + y_2}{2}\right)$
Solve for $x_2$ : Solve for $x_2$ : $-3 = \frac{9 + x_2}{2}$ $-6 = 9 + x_2$ $x_2 = -6 - 9$ $x_2 = -15$
Solve for $y_2$ : Solve for $y_2$ : $-16 = \frac{(8 + y_2)}{2}$ $-32 = 8 + y_2$ $y_2 = -32 - 8$ $y_2 = -40$

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