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ABC is an isosceles right triangle in which AB has a slope of -1 and mABC = 90°. ABC is dilated by a scale factor of 1.8 with the origin as the center of dilation, resulting in the image A'B'C'. What is the slope of B'C'?

A) - 1
B) 0
C) 1
D) 1.5
E) 2

Respuesta :

Answer:

C) 1

Step-by-step explanation:

Dilation is a transformation which produces an image that is the same shape as the original, but is a different size.

Here,

ABC is dilated by a scale factor of 1.8 with the origin as the center of dilation, resulting in the image A'B'C',

⇒ [tex]\triangle ABC\sim \triangle A'B'C'[/tex]

⇒ [tex]m\angle ABC = m\angle A'B'C'[/tex]

[tex]m\angle ABC = 90^{\circ}[/tex] ( Given )

[tex]\implies m\angle A'B'C' = 90^{\circ}[/tex]

⇒ AB ⊥ BC

Now, the slope of AB = - 1

By the property of perpendicular line segments,

Slope of AB × Slope of BC = - 1

⇒ - 1 × Slope of BC = - 1

⇒ Slope of BC = 1

Now, By the property of dilation,

BC ║ B'C'

⇒ Slope of B'C' = Slope of BC

⇒ Slope of B'C' = 1

Option C is correct.

lublana

Answer:

C)1

Step-by-step explanation:

We are given that triangle ABC is an isosceles right triangle.

Slope of AB=-1

[tex]m\angle ABC=90^{\circ}[/tex]

Triangle A'B'C' is the image of triangle ABC after dilation by scale factor 1.8 with the origin as the center of dilation.

We know that dilation is that transformation in which shape does not change of given figure but size changes after dilation.

Therefore, triangle A'B'C' is a isosceles right triangle.

[tex]\triangle ABC\sim \triangle A'B'C'[/tex]

We have to find the slope B'C'

AB is perpendicular to BC.

When two lines are perpendicular

Slope of one line=-[tex]\frac{1}{slope\;of\;another\;line}[/tex]

Therefore,

Slope of BC=[tex]\frac{-1}{slope\;of\;AB}=\frac{-1}{-1}=1[/tex]

Side BC is parallel to B'C'

When two lines are parallel then slope of lines are equal.

Slope of BC=Slope Of B'C'

Slope of B'C=1

Answer:1

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