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OlgaM077 [116]
3 years ago
14

(10 points) Solve for x. Assume that the lines which appear to be tangent are tangent

Mathematics
1 answer:
Ganezh [65]3 years ago
8 0

The two tangent lines for each circle are the same length. Set the equations to equal and solve for x.

1. 4x + 3 = 3x +12

Subtract 3x from each side:

x +3 = 12

Subtract 3 from both sides:

x = 9

2. -2 + x = 2x -7

Subtract 2x from both sides:

-2 - x =  -7

Add 2 to each side:

-x = -5

Multiply both sides by -1:

x = 5

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Step-by-step explanation:

From the picture attached,

9). When pre-image KLMN is a reflected across x = -2, we get image K'L'M'N'.

    Therefore, line of reflection will be x = -2

10). In this picture line of reflection which passes through origin (0, 0) and slope = 1 will be,

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11). Since, all the points of image and preimage points are equidistant from x-axis, line of reflection will be x-axis or y = 0.

12). Line passing through midpoint of the line joining BB' will be the line of reflection.

Coordinates of B → (5, 3)

Coordinates of B' → (5, 7)

Coordinates of BB' → (\frac{5+5}{2},\frac{3+7}{2}) → (5, 5)

Therefore, line of reflection will be y = 5.

13). Coordinates of G → (2, 5)

     Coordinates of G' → (-2, 5)

     Midpoint joining the segment GG' → (\frac{2-2}{2},\frac{5+5}{2}) → (0, 5)

     Therefore, line of reflection → x = 0

14). Line of reflection passing through (0, 0) and slope = -1

     y = -x

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