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3 years ago

1. R(2,7) Across the y-axis Across the x-axis:

Mathematics

2 answers:

AnnyKZ [126]3 years ago

6 0

Answer:

(- 2, 7 ) and (2, - 7 )

Step-by-step explanation:

Under a reflection in the y- axis

a point (x, y ) → (- x, y ) , then

(2, 7 ) → (- 2, 7 )

Under a reflection in the x- axis

a point (x, y ) → (x, - y ) , then

(2, 7 ) → (2, - 7 )

Ivan3 years ago

4 0

Answer:

Point R(2,7) has coordinates of (x, y), where x = 2 and y = 7, so

Across the y-axis = 7
Across the x-axis = 2

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100 points! Mhanifa can you please help? Look at the picture attached. I will mark brainliest!

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Answer:

1 and 2.

Midpoints calculated, plotted and connected to make the triangle DEF, see the attached.

D= (-2, 2), E = (-1, -2), F = (-4, -1)

As per definition, midsegment is parallel to a side.

Parallel lines have same slope.

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Find the slopes of FE and AB and compare.

m(FE) = (-2 - (- 1))/(-1 - (-4)) = -1/3
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Find the slopes of DE and AC and compare.

m(DE) = (-2 - 2)/(-1 - (-2)) = -4/1 = -4
m(AC) = (-5 - 3)/(-3 - (-5)) = -8/2 = -4
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As per definition, midsegment is half the parallel side.

We'll show that FD = 1/2CB

FD = $\sqrt{(2+1)^2+(-2+4)^2}$ = $\sqrt{3^2+2^2}$ = $\sqrt{13}$
CB = $\sqrt{(1 + 5)^2+(1+3)^2}$ = $\sqrt{6^2+4^2}$ = 2 $\sqrt{13}$
As we see FD = 1/2CB

FE = 1/2AB

FE = $\sqrt{(-4+1)^2+(-1+2)^2}$ = $\sqrt{3^2+1^2}$ = $\sqrt{10}$
AB = $\sqrt{(-5 -1)^2+(3-1)^2}$ = $\sqrt{6^2+2^2}$ = 2 $\sqrt{10}$
As we see FE = 1/2AB

DE = 1/2AC

DE = $\sqrt{(-2+1)^2+(2+2)^2}$ = $\sqrt{1^2+4^2}$ = $\sqrt{17}$
AC = $\sqrt{(-5 +3)^2+(3+5)^2}$ = $\sqrt{2^2+8^2}$ = 2 $\sqrt{17}$
As we see DE = 1/2AC

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1. R(2,7) Across the y-axis Across the x-axis:​

1. R(2,7) Across the y-axis Across the x-axis: