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3 years ago

7

Pls help me ill give brailiest do very soon

1 answer:

Nikitich [7]3 years ago

5 0

A.6x+34
B.8t+38
C.2y-14
D.11n+16

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What point Can be found by taking the average of the endpoints of a line segment?

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Step-by-step explanation:

This operation would result in the "midpoint" of the line segment.

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2 years ago

In the coordinate plane, the point A, 4− 2 is translated to the point A′, 1− 5. Under the same translation, the points B, 7− 4 a

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Answer:

The coordinates of B' and C' are $B'(x,y) = (4, -7)$ and $C'(x,y) = (-1, -8)$ , respectively.

Step-by-step explanation:

From the Linear Algebra, we define the translation of a given point as:

$O'(x,y) = O(x,y) + T(x,y)$ (1)

Where:

$O(x,y)$ - Original point, dimensionless.

$T(x,y)$ - Translation vector, dimensionless.

$O'(x,y)$ - Translated point, dimensionless.

If we know that $A'(x,y) = (1, -5)$ and $A(x,y) = (4,-2)$ , then the translation vector is:

$T(x,y) = A'(x,y)-A(x,y)$ (2)

$T(x,y) = (1,-5)-(4,-2)$

$T(x,y) = (-3,-3)$

If we know that $B(x,y) = (7,-4)$ , $C(x,y) = (2,-5)$ and $T(x,y) = (-3,-3)$ , then the translated points are, respectively:

$B'(x,y) = B(x,y)+T(x,y)$ (3)

$B'(x,y) = (7,-4) +(-3,-3)$

$B'(x,y) = (4, -7)$

$C'(x,y) = C(x,y) +T(x,y)$

$C'(x,y) = (2,-5) + (-3,-3)$

$C'(x,y) = (-1, -8)$

The coordinates of B' and C' are $B'(x,y) = (4, -7)$ and $C'(x,y) = (-1, -8)$ , respectively.

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