Reflecting a point

\(\displaystyle (x,y)\)

over the y-axis geometrically is the same as negating the x-coordinate of the ordered pair to obtain

\(\displaystyle (-x,y)\).

Thus, since our initial point was

\(\displaystyle (2,-2)\) 

and we want to reflect it over the y-axis, we obtain the reflection by negating the first term of the ordered pair to get

\(\displaystyle (-2,-2)\).

A line of symmetry of a figure is about which the reflection of the figure is the figure itself. The diagram below shows the only line of symmetry of the figure.

The correct response is one.

The reflection of the point at \(\displaystyle (x, y)\) about the \(\displaystyle x\)-axis is the point at \(\displaystyle (x, -y)\); therefore, the image of \(\displaystyle A (7, 4)\) is \(\displaystyle A' (7, -4)\). Since these two points have the same \(\displaystyle x\)-coordinate, the distance between them is the absolute value of the difference between their \(\displaystyle y\)-coordinates:

\(\displaystyle AA' = |4 - (-4)| = |8| = 8\)

Refer to the figure below, which shows the reflection of the given hexagon about \(\displaystyle \overleftrightarrow{AD}\) ; we will call \(\displaystyle A '\) the image of \(\displaystyle A\), call \(\displaystyle B'\) the image of \(\displaystyle B\), and so forth.

Now, refer to the figure below, which shows the reflection of the image about \(\displaystyle \overleftrightarrow{MN}\) ; we will call \(\displaystyle A ''\) the image of \(\displaystyle A'\), call \(\displaystyle B' '\) the image of \(\displaystyle B'\), and so forth.

Note that the vertices coincide with those of the original hexagon, and that the images of the points are in the same clockwise order as the original points. Since \(\displaystyle A ''\) coincides with \(\displaystyle B\), \(\displaystyle B''\) coincides with \(\displaystyle C\), and so forth, a clockwise rotation of five-sixth of a complete turn - that is, \(\displaystyle \frac{5}{6} \times 360 ^{\circ } = 300 ^{\circ }\),

is required to make each point its own image under the three transformations.

Refer to the figure below, which shows the reflection of \(\displaystyle A\) about \(\displaystyle \overleftrightarrow{BE}\) ; we will call this image \(\displaystyle A ''\).

Note that \(\displaystyle A ''\) coincides with \(\displaystyle C\). Now, refer to the figure below, which shows the reflection of \(\displaystyle A ''\) about \(\displaystyle \overleftrightarrow{MN}\); we will call this image - the final image - \(\displaystyle A'\)

Note that \(\displaystyle A'\) coincides with \(\displaystyle F\), making this the correct response.

A dilation is the stretching or shrinking of a figure.

A rotation is the turning of a a figure about a point. 

A reflection is the flipping of a figure across a line.

A translation is is the sliding of a figure in a direction. 

With a translation, the image is not only congruent to its original size and shape, but its orientation remains the same. A translation fits this figure best because the shape seems to move upward and rightward without changing size, shape, or orientation.

If the graph of an equation is translated to the right \(\displaystyle h\) units, and upward \(\displaystyle k\) units, the equation of the image can be found by replacing \(\displaystyle x\) with \(\displaystyle x-h\) and \(\displaystyle y\) with \(\displaystyle y-k\) in the equation of the original graph. 

Since we are moving the graph of the equation

\(\displaystyle 4x- 7y = 15\)

left four units and down six units, we set \(\displaystyle h = -4\) and \(\displaystyle k = -6\); we can replace \(\displaystyle x\) with \(\displaystyle x - (-4)\), or \(\displaystyle x+ 4\), and \(\displaystyle y\) with \(\displaystyle y - (-6 )\), or \(\displaystyle y + 6\). The equation of the image can be written as

\(\displaystyle 4 (x+ 4)- 7 (y+6) = 15\)

Simplify by distributing:

\(\displaystyle 4 x+4( 4)- 7 y-7(6) = 15\)

\(\displaystyle 4 x+16- 7 y-42 = 15\)

Collect like terms:

\(\displaystyle 4 x- 7 y +16-42 = 15\)

\(\displaystyle 4 x- 7 y -26 = 15\)

Add 26 to both sides:

\(\displaystyle 4 x- 7 y -26 +26 = 15+26\)

\(\displaystyle 4 x- 7 y = 41\),

the correct choice.

If the graph of an equation is translated to the right \(\displaystyle h\) units, and upward \(\displaystyle k\) units, the equation of the image can be found by replacing \(\displaystyle x\) with \(\displaystyle x-h\) and \(\displaystyle y\) with \(\displaystyle y-k\) in the equation of the original graph. 

Since we are moving the graph of the equation

\(\displaystyle x^{2}+ y^{2}= 100\)

right two units and up five units, we set \(\displaystyle h = 2\) and \(\displaystyle k = 5\); we can therefore replace \(\displaystyle x\) with \(\displaystyle x- 2\) and \(\displaystyle y\) with \(\displaystyle y - 5\). The equation of the image can be written as 

\(\displaystyle (x-2)^{2}+( y-5)^{2}= 100\)

This can be rewritten by applying the binomial square pattern as follows:

\(\displaystyle x^{2}- 2 (x) (2)+ 2^{2}+ y^{2}- 2 (5)(x)+ 5^{2} = 100\)

\(\displaystyle x^{2}- 4x+4+ y^{2}-10y+2 5 = 100\)

Collect like terms; the equation becomes

\(\displaystyle x^{2}- 4x + y^{2}-10y+29 = 100\)

Subtract 100 from both sides:

\(\displaystyle x^{2}- 4x + y^{2}-10y+29- 100 = 100 - 100\)

\(\displaystyle x^{2}- 4x + y^{2}-10y-71=0\)

If the graph of an equation is translated to the right \(\displaystyle h\) units, and upward \(\displaystyle k\) units, the equation of the image can be found by replacing \(\displaystyle x\) with \(\displaystyle x-h\) and \(\displaystyle y\) with \(\displaystyle y-k\) in the equation of the original graph. 

Since we are moving the graph of the equation

\(\displaystyle y = x^{2}+ 5x - 7\)

right four units and down two units, we set \(\displaystyle h = 4\) and \(\displaystyle k = -2\); we can therefore replace \(\displaystyle x\) with \(\displaystyle x- 4\), and \(\displaystyle y\) with \(\displaystyle y - (-2)\), or \(\displaystyle y+ 2\). The equation of the image can be written as

\(\displaystyle y+2 = (x-4)^{2}+ 5(x-4) - 7\)

The expression at right can be simplified. First, use the distributive property on the middle expression:

\(\displaystyle y+2 = (x-4)^{2}+ 5x-5 (4) - 7\)

\(\displaystyle y+2 = (x-4)^{2}+ 5x-20 - 7\)

Now, simplify the first expression by using the binomial square pattern:

\(\displaystyle y+2 = x^{2} -2 (x)(4) +4^{2}+ 5x-20 - 7\)

\(\displaystyle y+2 = x^{2} -8x+16+ 5x-20 - 7\)

Collect like terms on the right:

\(\displaystyle y+2 = x^{2} -8x+ 5x+16 -20 - 7\)

\(\displaystyle y+2 = x^{2} -3x-11\)

Subtract 2 from both sides:

\(\displaystyle y+2 -2= x^{2} -3x-11-2\)

\(\displaystyle y= x^{2} -3x-13\),

the equation of the image.

If the graph of an equation is translated to the right \(\displaystyle h\) units, and upward \(\displaystyle k\) units, the equation of the image can be found by replacing \(\displaystyle x\) with \(\displaystyle x-h\) and \(\displaystyle y\) with \(\displaystyle y-k\) in the equation of the original graph. 

Since we are moving the graph of the equation

\(\displaystyle y = | 3x - 8 |\)

left three units and down five units, we set \(\displaystyle h = -3\) and \(\displaystyle k = -5\); we can therefore replace \(\displaystyle x\) with \(\displaystyle x- (-3)\), or \(\displaystyle x +3\), and \(\displaystyle y\) with \(\displaystyle y - (-5)\), or \(\displaystyle y+ 5\). The equation of the image can be written as

\(\displaystyle y+5 = | 3(x+3) - 8 |\)

We can simplify the expression on the right by distributing:

\(\displaystyle y+5 = | 3x+3 \cdot 3 - 8 |\)

\(\displaystyle y+5 = | 3x+9- 8 |\)

Collect like terms:

\(\displaystyle y+5 = | 3x+1 |\)

Subtract 5 from both sides:

\(\displaystyle y+5 -5 = | 3x+1 | - 5\)

\(\displaystyle y = | 3x+1 | - 5\),

the correct equation of the image.