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daser333 [38]
3 years ago
5

What is the reflection of point P(−1, 6) across the line y = x?

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
6 0

ANSWER

P'(6,-1)

EXPLANATION

The mapping for a reflection in the line y=x is

The line y=x is the mirror line for a function and it's inverse.

(x,y)\to (y,x)

The x and y coordinates swap position.

The reflection of point P(−1, 6) across the line y = x is

P'(6,-1)

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Consider the graph of the line y = x – 4 and the point (−4, 2).
Hitman42 [59]

Answer:

a. The slope  of a line parallel to the given line is 1

b. A point on the line parallel to the given line, passing through (−4, 2), is  (1,7)

c. The slope of the line perpendicular to the given line is -1

d. A point on the line perpendicular to the given line, passing through (−4, 2), is (3,-5)

Step-by-step explanation:

The equation of the line in Slope-intercept form  is:

y=mx+b

Where m is the slope and b is the y-intercept.

a. For the line y = x - 4

You can identify that:

m=1

 By definition, two lines are parallel if they have the same slope. Then, the  slope of a line parallel to the given line is:

m=1

b. The equation of the line in Point-slope form is:

y -y_1 = m(x - x_1)

Where m is the slope and (x_1,y_1)  is a point of the line.

Given the point (-4,2), substitute this point and the slope of the line into the equation:

 y -2 = (x +4)

Give a value to "x", substitute it into this equation and solve for "y":

For x=1 :

y -2 = (1 +4)

y= 5+2

y= 7

Then, you get the point (1,7)

c. The slopes of perpendicular lines are negative reciprocals, then the  slope of a line perpendicular to the given line is:

m=-\frac{1}{1}\\\\m=-1

d. Given the point (-4,2), substitute this point and the slope of the line into the equation:

 y -2 = -1(x +4)

 y -2 = -(x +4)

Give a value to "x", substitute it into this equation and solve for "y":

For x=3 :

y -2 = -(3 +4)

y= -7+2

y= -5

Then, you get the point (3,-5)

8 0
3 years ago
Use trigonometric identities to solve each equation within the given domain.
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Recall that the tangent function is defined by

tan(<em>x</em>) = sin(<em>x</em>)/cos(<em>x</em>)

Also recall the double angle identity for sine,

sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)

Then the equation is the same as

3 sin(<em>x</em>)/cos(<em>x</em>) = 4 sin(<em>x</em>) cos(<em>x</em>)

Move everything to one side to prepare to factorize:

3 sin(<em>x</em>)/cos(<em>x</em>) - 4 sin(<em>x</em>) cos(<em>x</em>) = 0

sin(<em>x</em>)/cos(<em>x</em>) (3 - 4 cos²(<em>x</em>)) = 0

As long as cos(<em>x</em>) ≠ 0, we can omit the term in the denominator, so we're left with

sin(<em>x</em>) (3 - 4 cos²(<em>x</em>)) = 0

and so

sin(<em>x</em>) = 0   <u>or</u>   3 - 4 cos²(<em>x</em>) = 0

sin(<em>x</em>) = 0   <u>or</u>   cos²(<em>x</em>) = 3/4

sin(<em>x</em>) = 0   <u>or</u>   cos(<em>x</em>) = ±√3/2

On the interval [0, 2<em>π</em>),

• sin(<em>x</em>) = 0 for <em>x</em> = 0 and <em>x</em> = <em>π</em>

• cos(<em>x</em>) = √3/2 for <em>x</em> = <em>π</em>/6 and <em>x</em> = 11<em>π</em>/6

• cos(<em>x</em>) = -√3/2 for <em>x</em> = 5<em>π</em>/6 and <em>x</em> = 7<em>π</em>/6

(None of these <em>x</em> make cos(<em>x</em>) = 0, so we don't have to omit any extraneous solutions.)

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Answer:

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Step-by-step explanation:

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Free_Kalibri [48]

Answer:

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Step-by-step explanation:

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