Question

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

A trapezoid ABCD drawn on a 4−quadrant coordinate grid with vertices A(−8, 5), B(−5, 5), C(−6, 7) and D(−7, 7). Point J located at (−6, −8) and point M located at (−3, −2)

The badge is enlarged and plotted on the coordinate grid as figure JKLM with point J located at (−6, −8) and point M located at (−3, −2).

Which of these could be the coordinates for point L?

(−3, −8)
(−1, −2)
(2, −8)
(0, −2)

Answer 1

D- (0,2)

Explanation:

i took the test and this was the answer

Answer 2

The correct option is 4.

Explanation:

It is given vertices are A(−8, 5), B(−5, 5), C(−6, 7) and D(−7, 7). The figure ABCD enlarged and plotted on the coordinate grid as figure JKLM. The vertices of image are J(-6,-8) and M(-3,-2).

The y-coordinate of C and D are same, therefore the y-coordinate of L and M must be same.

Let the coordinate of L be (x,-2).

Distance formula:

`D=√((x_2-x_1)^2+(y_2-y_1)^2)`

`AD=√((-7-(-8))^2+(7-5)^2)=\srqt{5}`

`JM=√((-3-(-6))^2+(-2-(-8))^2)=\srqt{45}=3\srqt{5}`

Scale factor is

`(JM)/(AD)=(3√(5))/(√(5))=3`

Therefore the scale factor is 3.

The distance between CD is

`CD=√((-7-(-6))^2+(7-7)^2)=1`

Since the scale factor is 3, therefore the LM is 3 units.

`LM=√((x-(-3))^2+(-2-(-2))^2)`

`3=x+3`

`x=0`

The coordinates of LM are (0,-2).

Therefore option 4 is correct.