Question

(b) part c
which transformations of triangle ( abc ) would result in the image of the triangle
being, located only in the first quadrant of the coordinate plane?
select each correct answer.
a a translation right 3 units
b a reflection across ( x = 4 )
c a reflection across ( y = -x )
d a rotation of ( 90^{circ} ) counterclockwise about vertex ( a ).
e a reflection across ( x = -2 ), then a translation right 9 units
f a reflection across the ( x )-axis, then a translation up 13 units
(c) part c
triangle ( abc ) will be reflected across the ( x )-axis and then reflected across the ( y )-axis
to produce ( a^{prime prime} b^{prime prime} c^{prime prime} ). given ( (x, y) ) describes a vertex of ( a ), which
coordinates describe the corresponding vertex of ( a^{prime prime} )?
(a) ( (x - 1, y - 1) )
(b) ( (-x,-y) )
(c) ( (x,-y) )
(d) ( (-x, y) )

Explanation:

Part (b)

Step1: Analyze translation right 3 units

A translation right 3 units will keep the triangle in the first quadrant if it was originally there. Since moving right only increases the \(x -\)coordinate.

Step2: Analyze reflection across \(x = 4\)

Let the vertices of \(\triangle ABC\) have \(x\) - coordinates \(x_1,x_2,x_3\). After reflection across \(x = 4\), the new \(x\) - coordinates are \(x'=8 - x\). If the original \(x\) - coordinates are such that \(8 - x>0\) (for example, if original \(x\) - coordinates are less than 8), the image will be in the first quadrant.

Step3: Analyze reflection across \(y=-x\)

The transformation \( (x,y)\to(-y,-x)\). If \((x,y)\) is in the first quadrant (\(x>0,y > 0\)), then \((-y,-x)\) is in the third quadrant.

Step4: Analyze rotation of \(90^{\circ}\) counter - clockwise about vertex \(A\)

A rotation about a vertex in the first quadrant will keep the triangle in the first quadrant.

Step5: Analyze reflection across \(x=-2\) then translation up 13 units

Reflection across \(x =- 2\) gives \(x'=-4 - x\). If \(x>0\), then \(x'=-4 - x<0\). After translation up 13 units (\(y'=y + 13\)), the \(x\) - coordinate is negative, so the triangle will not be in the first quadrant.

Step6: Analyze reflection across the \(x -\)axis then translation up 13 units

Reflection across the \(x -\)axis gives \((x,-y)\). If \(y>0\), then \(-y<0\). After translation up 13 units (\(y'=-y + 13\)), if \(y<13\), \(y'>0\), but the \(x\) - coordinate remains the same. If the original \(x>0\), the image is in the first quadrant.

Step1: Reflection across the \(x -\)axis

The rule for reflection across the \(x -\)axis is \((x,y)\to(x,-y)\).

Step2: Reflection across the \(y -\)axis

The rule for reflection across the \(y -\)axis is \((x,y)\to(-x,y)\).

Step3: Combine the two reflections

First, reflect \((x,y)\) across the \(x -\)axis to get \((x,-y)\), then reflect \((x,-y)\) across the \(y -\)axis. Using the rule \((x,y)\to(-x,y)\) for reflection across the \(y -\)axis, we substitute \(y\) with \(-y\) and get \((-x,-y)\)

Answer:

A. a translation right 3 units; B. a reflection across \(x = 4\); D. a rotation of \(90^{\circ}\) counterclockwise about vertex \(A\); F. a reflection across the \(x -\)axis, then a translation up 13 units

Part (c)