Which points are separated by a distance of 4 units? a. (7,7) (1,7) b. (3,3) (1,3) c. (4,2) (8,2) d. (2,4) (5,4)

Answer:

c

Step-by-step explanation:

with the knowledge that we are comparing the x position we can get the equation |x_1-x_2|=4, then just do the maths

(4,2) (8,2) points are separated by a distance of 4 units.

The 2D distance formula gives the shortest distance between two points in a two-dimensional plane.

Distance Formula =  [tex]\sqrt{(x_{2} -x_{1} )^{2} +(y_{2}-y_{1} )^{2} }[/tex]

Given

Distance of 4 units

Option A:

[tex](x_{1} ,y_{1})=(7,7)[/tex]

[tex](x_{2} ,y_{2})=(1,7)[/tex]

Distance = [tex]\sqrt{(x_{2} -x_{1} )^{2} +(y_{2}-y_{1} )^{2} }[/tex]

= [tex]\sqrt{(1-7)^{2}+(7-7)^{2} } =6[/tex]

Option B:

[tex](x_{1} ,y_{1})=(3,3)[/tex]

[tex](x_{2} ,y_{2})=(1,3)[/tex]

Distance = [tex]\sqrt{(x_{2} -x_{1} )^{2} +(y_{2}-y_{1} )^{2} }[/tex]

= [tex]\sqrt{(1-3)^{2}+(3-3)^{2} } =2[/tex]

Option C:

[tex](x_{1} ,y_{1})=(4,2)[/tex]

[tex](x_{2} ,y_{2})=(8,2)[/tex]

Distance = [tex]\sqrt{(x_{2} -x_{1} )^{2} +(y_{2}-y_{1} )^{2} }[/tex]

= [tex]\sqrt{(8-4)^{2}+(2-2)^{2} } =4[/tex]

Option C is correct.

Option D:

[tex](x_{1} ,y_{1})=(2,4)[/tex]

[tex](x_{2} ,y_{2})=(5,4)[/tex]

Distance = [tex]\sqrt{(x_{2} -x_{1} )^{2} +(y_{2}-y_{1} )^{2} }[/tex]

= [tex]\sqrt{(5-2)^{2}+(4-4)^{2} } =3[/tex]

Hence, (4,2) (8,2) points are separated by a distance of 4 units.

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