Answer:
B) (21, -14) and (21, -22)
Step-by-step explanation:
Given,
The area of the square = 64 square unit,
Let x be the side of the square,
[tex]\implies x^2 = 64[/tex]
[tex]\implies x = 8[/tex]
Hence, the side of the square = 8 unit,
⇒ The distance between the adjacent vertices in the square = 8 units,
By the distance formula,
The distance between (3, -35) and (3, -28) is,
[tex]\sqrt{(3-3)^2+(-28-(-35))^2[/tex]
[tex]=\sqrt{0+(-28+35)^2}[/tex]
[tex]=\sqrt{7^2}[/tex]
[tex]=7\neq 8[/tex]
Thus, (3, -35) and (3, -28) can not be the vertices of the square.
The distance between (21, -14) and (21, -22) is,
[tex]\sqrt{(21-21)^2+(-22-(-14))^2[/tex]
[tex]=\sqrt{0+(-22+14)^2}[/tex]
[tex]=\sqrt{8^2}[/tex]
[tex]=8[/tex]
Thus, (21, -14) and (21, -22) are the vertices of the square.
The distance between (32, -42) and (32, -7) is,
[tex]\sqrt{(32-32)^2+(-7-(-42))^2[/tex]
[tex]=\sqrt{0+(-7+42)^2}[/tex]
[tex]=\sqrt{35^2}[/tex]
[tex]=35\neq 8[/tex]
Thus, (32, -42) and (32, -7) can not be the vertices of the square.
The distance between (74, 19) and (82, 27) is,
[tex]\sqrt{(82-74)^2+(27-19)^2[/tex]
[tex]=\sqrt{8^2+8^2}[/tex]
[tex]=\sqrt{64+64}[/tex]
[tex]=\sqrt{128}[/tex]
[tex]=8\sqrt{2}\neq 8[/tex]
Thus, (74, 19) and (82, 27) can not be the vertices of the square.
