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Bas_tet [7]
3 years ago
15

What is the equation of a line that is parallel to the line y = 5x – 1 and passes

Mathematics
1 answer:
dsp733 years ago
5 0

Answer:

y = 5x - 23

Step-by-step explanation:

Since that is ll to y = 5x - 1, slope must be same as that of y = 5y - 1.

Compare y = 5y - 1 with y = mx + c, where m is slope.

Slope of the required line = 5.

And a point lying on that line is (4,-3). Therefore,

Equation is:

=> y - (-3) = 5(x - 4)

=> y + 3 = 5x - 20

=> y = 5x - 23

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Answer:

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2 years ago
Segment FG begins at point F(-2, 4) and ends at point G (-2, -3). Segment FG is translated by (x, y) → (x – 3, y + 2) and then r
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Answer:

The length of the segment F'G' is 7.

Step-by-step explanation:

From Linear Algebra we define reflection across the y-axis as follows:

(x',y')=(-x, y), \forall\, x, y\in \mathbb{R} (Eq. 1)

In addition, we get this translation formula from the statement of the problem:

(x',y') =(x-3,y+2), \forall \,x,y\in \mathbb{R} (Eq. 2)

Where:

(x, y) - Original point, dimensionless.

(x', y') - Transformed point, dimensionless.

If we know that F(x,y) = (-2, 4) and G(x,y) = (-2,-3), then we proceed to make all needed operations:

Translation

F''(x,y) = (-2-3,4+2)

F''(x,y) = (-5,6)

G''(x,y) = (-2-3,-3+2)

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F'(x,y) = (5, 6)

G'(x,y) = (5,-1)

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F'G' = \sqrt{(5-5)^{2}+[(-1)-6]^{2}}

F'G' = 7

The length of the segment F'G' is 7.

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