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Mathematics 56 Online
D14610:

Triangle XYZ is shown on the coordinate plane. If triangle XYZ is translated using the rule (x, y) → (x + 1, y − 4) and then reflected across the y-axis to create triangle X″Y″Z″, what is the location of Z″?

D14610:

1 attachment
somnium:

i gotchu gimme one sec

somnium:

Z' will lie on (-2,-1)

somnium:

to do this, a translation is basically scooting the shape

somnium:

so since the rule is (x,y), (x+1, y-4), you have to add 1 to the initial x value, and remove 4 from the y

somnium:

after doing that, Z lies on (2,-1)

somnium:

then you have to reflect across the y-axis

somnium:

when you reflect, you basically fold the shape over the axis. mathematically, if you flip across the y-axis, the x-value is negated, and if you flip across the x-axis, then the y-value is negated

somnium:

(2,-1) is then flipped, so you have to negate the x-value (which is 2), so then your final prime vertex Z is on the point (-2,-1). i hope this helps, and if you need a more detailed explanation, feel free to ask!!

D14610:

i think i understand, if you wanna answer 1 more for me i'd appreciate it greatly

somnium:

@d14610 wrote:
i think i understand, if you wanna answer 1 more for me i'd appreciate it greatly
sure, more than happy to!

D14610:

@somnium wrote:
@d14610 wrote:
i think i understand, if you wanna answer 1 more for me i'd appreciate it greatly
sure, more than happy to!

1 attachment
somnium:

@d14610 wrote:
@somnium wrote:
@d14610 wrote:
i think i understand, if you wanna answer 1 more for me i'd appreciate it greatly
sure, more than happy to!
what are the options on this one?

D14610:

@somnium wrote:
@d14610 wrote:
@somnium wrote:
@d14610 wrote:
i think i understand, if you wanna answer 1 more for me i'd appreciate it greatly
sure, more than happy to!
what are the options on this one?
x‒axis, y = x, x‒axis, y = x y-axis, x‒axis, y-axis x‒axis, y-axis, y-axis y = x, x‒axis, y = x, y-axis

somnium:

@d14610 wrote:
@somnium wrote:
@d14610 wrote:
@somnium wrote:
@d14610 wrote:
i think i understand, if you wanna answer 1 more for me i'd appreciate it greatly
sure, more than happy to!
what are the options on this one?
x‒axis, y = x, x‒axis, y = x y-axis, x‒axis, y-axis x‒axis, y-axis, y-axis y = x, x‒axis, y = x, y-axis
then the answer will be a, x‒axis, y = x, x‒axis, y = x. this one is a bit more complicated, but the answer is much simpler. basically the question is asking how you reflect it to move it, without moving the end product. since the first reflection is the same as the second, it is performing the same action again, so it lands on its original points.

D14610:

thank you so so much

somnium:

@d14610 wrote:
thank you so so much
of course, i hope that that helps you to understand the process ^^

D14610:

yeah it's those kinds of questions that really irritate me it's so tedious

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