Question

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 1

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

Answer 2

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

What is distance formula?

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

Distance Formula =  $\sqrt{(x_{2} -x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

Given

Distance of 4 units

Option A:

$(x_{1} ,y_{1})=(7,7)$

$(x_{2} ,y_{2})=(1,7)$

Distance = $\sqrt{(x_{2} -x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

= $\sqrt{(1-7)^{2}+(7-7)^{2} } =6$

Option B:

$(x_{1} ,y_{1})=(3,3)$

$(x_{2} ,y_{2})=(1,3)$

Distance = $\sqrt{(x_{2} -x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

= $\sqrt{(1-3)^{2}+(3-3)^{2} } =2$

Option C:

$(x_{1} ,y_{1})=(4,2)$

$(x_{2} ,y_{2})=(8,2)$

Distance = $\sqrt{(x_{2} -x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

= $\sqrt{(8-4)^{2}+(2-2)^{2} } =4$

Option C is correct.

Option D:

$(x_{1} ,y_{1})=(2,4)$

$(x_{2} ,y_{2})=(5,4)$

Distance = $\sqrt{(x_{2} -x_{1} )^{2} +(y_{2}-y_{1} )^{2} }$

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

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

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