Question

Help Please? :)

An artist designed a badge for a school club. Figure ABCD on the coordinate grid below shows the shape of the badge:

The badge is enlarged and plotted on the coordinate grid as figure JKLMwith point J located at (−6, −8) and point M located at (−3, −2). Which of these could be the coordinates for point L?
Select one:
a. (−3, −8)
b. (−1, −2)
c. (2, −8)
d. (0, −2)

Answer 1

The answer would be (0, -2).  You can see from the graph that the original slope is 2.  So if you use the same slope on the new grid, you see that is expanded by a factor of 3.  Following ABCD counter clockwise around the original grid, you see that on the new grid JKLM, L would fall on the point C falls in the original grid.  Count over from D to C on original grid (1 unit).  Count on new grid from M to L and it should be 3 units over (because we expanded by a factor of 3.  That's point (0, -2).

Answer 2

Answer:

The correct option is d.

Step-by-step explanation:

It is given that the figure JKLM is image of ABCD. So the side MJ is the image of DA.

From the figure it is noticed that the A(-8,5) and D(-7,7).

Distance formula is

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AD=\sqrt{(-7-(-8))^2+(7-5)^2}$

$AD=\sqrt{1+4}$

$AD=\sqrt{5}$

The points of image are J(-6,-8) and M(-3,-2).

$JM=\sqrt{(-3-(-6))^2+(-2-(-8))^2}$

$JM=\sqrt{(3)^2+(6)^2}$

$JM=\sqrt{9+36}$

$JM=\sqrt{45}$

$JM=3\sqrt{5}$

The scale factor is,

$k=\frac{JM}{AD}=\frac{3\sqrt{5}}{\sqrt{5}} =3$

Therefore the corresponding sides of image are 3 times more than the preimage.

From the figure it is noticed that the distance between DC is 1 units.

Since the sides ML is image of DC, therefore the length of ML is 3 units.

The point C is 1 unit right from D, therefore the point L is 3 unit right from M. The y coordinate of M and L are same. To find the x coordinate of L we have to add 3 units in x coordinate of M.

$L=(-3+3,-2)$

$L=(0,-2)$

Therefore the option d is correct.