Triangle MNP is dilated according to the rule DO,1.5 (x,y)(1.5x, 1.5y) to create the image triangle M'N'P, which is not shown.



What are the coordinates of the endpoints of segment M'N'?
M'(-6, 9) and N'(4, 9)
M'(-6, 9) and N'(3, 9)
M'(-2, 3) and N'(7, 9)
M'(-2, 3) and N'(1, 3)

Triangle MNP is dilated according to the rule DO15 xy15x 15y to create the image triangle MNP which is not shown What are the coordinates of the endpoints of se class=

Respuesta :

Answer:

Option B is correct i.e.,  M'(-6,9) and N'(3,9).

Step-by-step explanation:

Given: ΔMNP mapped in graph ⇒ coordinates of vertices are M(-4,6), N(2,6) & P(-1,1)  and scale factor of dilation is 1.5 .i.e., 1.5(x,y) ⇒ (1.5x,1.5y)

ΔMNP dilated to ΔM'N'P'

To Find: Coordinates of M' &N'

since, scale factor of dilation is 1.5.

we get coordinates of M' by multiplying coordinates of M

⇒1.5 M(-4,6) = M'(1.5×(-4), 1.5×6)

                    =M' (-6,9)

similalrly coordinates of N' by multiplying coordinates of N

⇒1.5 N(2,6) = N'(1.5×2, 1.5×6)

                    =N' (3,9)

The coordinates of the endpoints of segment M'N' is M'(-6,9) and N'(3,9)

∴Option B is correct.

The correct option for the coordinates of end point of line segment is [tex]\fbox{\begin\\\ \bf option (b)\\\end{minispace}}[/tex].

Further Explanation:

Calculation:

The coordinate of the point [tex]P,M,N[/tex] can be obtained from the given Graph.

The coordinate of the point [tex]P[/tex] of [tex]\triangle\text{PMN}[/tex] is [tex](-1,1)[/tex].

The coordinate of the point [tex]N[/tex] of a triangle [tex]\triangle\text{PMN}[/tex] is [tex](2,6)[/tex].

The coordinates of the point [tex]M[/tex] of a triangle [tex]\triangle\text{PMN}[/tex] is [tex](-4,6)[/tex].

The image of triangle is created by the given rule,

[tex]\boxed{1.5(x,y)\rightarrow(1.5x,1.5y)}[/tex]

The new image of [tex]\triangle\text{PMN}[/tex] is [tex]\triangle\text{M'N'P'}[/tex].

The new coordinate of [tex]P[/tex] that is [tex]P'[/tex] is obtained as,

[tex]\boxed{\begin{aligned}1.5(x,y)&\rightarrow(1.5x,1.5y)\\1.5(-1,1)&\rightarrow(-1.5,1.5)\end{aligned}}[/tex]

The new coordinate of [tex]M[/tex] that is [tex]M'[/tex] is obtained as,

[tex]\boxed{\begin{aligned}1.5(x,y)&\rightarrow(1.5x,1.5y)\\1.5(-4,6)&\rightarrow(-6,9)\end{aligned}}[/tex]

The new coordinate of [tex]N[/tex] that is [tex]N'[/tex] is obtained as,

[tex]\boxed{\begin{aligned}1.5(x,y)&\rightarrow(1.5x,1.5y)\\1.5(2,6)&\rightarrow(3,9)\end{aligned}}[/tex]

Therefore, the coordinate for point [tex]M'[/tex] is [tex](-6,9)[/tex] and the coordinate for the point [tex]N'[/tex] is [tex](3,9)[/tex].

Thus, the correct option for the coordinates of end point of line segment is [tex]\fbox{\begin\\\ \bf option (b)\\\end{minispace}}[/tex].

Learn more:

1. A problem on triangle: https://brainly.com/question/7437053

2. A problem on transformation of triangle https://brainly.com/question/2992432

Answer details:

Grade: High school

Subject: Mathematics

Chapter: Triangle

Keywords: Equations, Triangle, dilate, Translate, image, size, x coordinate, y coordinate, shifted position endpoints, line segment, shifting, 1.5(x,y)=(1.5x,1.5y).