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Alexeev081 [22]
3 years ago
5

Choose all of the statements that correctly describe the transformation rule. Reflection over x-axis: (x, y) ? (?x, y) Reflectio

n over y-axis: (x, y) ? (x, ?y) Rotation of 90° counter-clockwise about origin: (x, y) ? (?y, x) Rotation of 180° counter-clockwise about origin: (x, y) ? (?x, ?y) Rotation of 270° counter-clockwise about origin: (x, y) ? (y, ?x)
Mathematics
1 answer:
matrenka [14]3 years ago
3 0
<h2>Answer:</h2>

Transformations are important subjects in geometry. In this exercise, these are the correct transformation rules:

<h3>1. Reflection over x-axis:</h3>

Consider the point (x,y), if you reflect this point across the x-axis you should multiply the y-coordinate by -1, so you get:

\boxed{(x,y)\rightarrow(x,-y)}

<h3>2. Reflection over y-axis: </h3>

Consider the point (x,y), if you reflect this point across the y-axis you should multiply the x-coordinate by -1, so you get:

\boxed{(x,y)\rightarrow(-x,y)}

<h3>3. Rotation of 90° counter-clockwise about origin: </h3>

Consider the point (x,y). To rotate this point by 90° around the origin in counterclockwise direction, you can always swap the x- and y-coordinates and then multiply the new x-coordinate by -1. In a mathematical language this is as follows:

\boxed{(x,y)\rightarrow(-y,x)}

<h3>4. Rotation of 180° counter-clockwise about origin:</h3>

Consider the point (x,y). To rotate this point by 180° around the origin, you can flip the sign of both the x- and y-coordinates. In a mathematical language this is as follows:

\boxed{(x,y)\rightarrow(-x,-y)}

<h3> 5. Rotation of 270° counter-clockwise about origin: </h3>

Rotate a point 270° counter-clockwise about origin is the same as rotating the point 90° in clock-wise direction. So the rule is:

\boxed{(x,y)\rightarrow(y,-x)}

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Write the equation of a line that passes through the points ( 2 , - 5 ) and ( 8 , - 2 )
kati45 [8]

Answer:

y=\frac{1}{2}x-6

Step-by-step explanation:

Hi there!

Linear equations are typically organized in slope-intercept form: y=mx+b where m is the slope and b is the y-intercept (the value of y when the line crosses the y-axis)

<u>1) Determine the slope (m)</u>

m=\frac{y_2-y_1}{x_2-x_1} where two points that fall on the line are (x_1,y_1) and (x_2,y_2)

Plug in the given points (2,-5) and (8,-2)

m=\frac{y_2-y_1}{x_2-x_1}\\=\frac{-2-(-5)}{8-2}\\=\frac{-2+5}{8-2}\\=\frac{3}{6}\\=\frac{1}{2}

Therefore, the slope of the line is \frac{1}{2}. Plug this into y=mx+b:

y=\frac{1}{2}x+b

<u>2) Determine the y-intercept (b)</u>

y=\frac{1}{2}x+b

Plug in one of the given points and solve for b

-5=\frac{1}{2}(2)+b\\-5=1+b

Subtract 1 from both sides to isolate b

-5-1=1+b-1\\-6=b

Therefore, the y-intercept of the line is -6. Plug this back into y=\frac{1}{2}x+b:

y=\frac{1}{2}x+(-6)\\y=\frac{1}{2}x-6

I hope this helps!

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3 years ago
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