Question

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

Answer 1

Answer:

Option B is right.

Step-by-step explanation:

Given that an artist designed a badge for a school club.

ABCD is given in the picture.

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

If JKLM is enlargement of ABCD, then scale factor would b the ratio of AD to JM

JM = $\sqrt{(5-2)^2+(4+2)^2} \\=\sqrt{45} \\3\sqrt{5}$

Scale factor = 3

AB = 3 hence JK = 9

DC = 1 and so ML = 3

ML is parallel to x axis with a distance of 3 units.

COordinates of L = (5+3, 4) = (8,4)

Option B is right

Answer 2

Answer:

(8, 4)

Step-by-step explanation:

AD is enlarged to JM

AD² = 2² + 1² = 4 + 1 =5

AD =  √5

then you plug in JM

JM² = 6² + 3² = 36 + 9 = 45

JM = √45 = 3 √5

so if

JM=3 AD

then

ML= 3 DC

then you add

5+3=8

the coordinates for point L are (8, 4)