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What is the distance between (3, 5.25) and (3, –8.75)? 6 units 8.25 units 11.75 units 14 units

User Bigredbob
by
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2 Answers

3 votes
By using distance formula :



\text{Distance formula,} \bold{ \boxed{ Distance = \sqrt{( x_(2) -x_(1))^(2)+(y_(2) - y_(1))^(2)) }}}





Given points = ( 3 , 5.25 ) and ( 3 , - 8.75 )



\bold{Taking \: \: \: x_(1)=3 \: \: , \: \: x_(2)= 3 \: \: , \: \: y_(1)= 5.25 \: \: , \: \: y_(2)= -8.75}




On applying formula, we get



Distance = \sqrt{ ( x_(2)-x_(1))^(2)+(y_(2)-y_(1))^2} \\ \\ \\ Distance = \sqrt{ ( 3 - 3 )^(2) + ( - 8.75 - 5.25 )^(2)} \\ \\ \\ Distance = \sqrt{ ( 0 )^(2) + ( - 14)^(2)} \\ \\ \\ Distance = \sqrt{ ( - 14 )^(2)} \\ \\ \\ Distance = \sqrt{ 14^(2)} \:\:\:\:\:\:\:\:\:\:\: \: \: \: \: \: \: \: \: | \bold{ ( - 14 )^(2) = 14^(2)} \\ \\ \\ Distance = {14}^{2 * (1)/(2) } \\ \\ \\ Distance = {14}^(1) \\ \\ \\ Distance = 14 \: units








Hence, Option D is correct.
User Kvadityaaz
by
8.3k points
3 votes

Answer:

D. 14 units.

Explanation:

We have been given coordinates of two points. We are asked to find the distance between both points.

We will distance formula to solve our given problem.


D=√((x_2-x_1)^2+(y_2-y_1)^2), where,


x_2-x_1 = Difference between two x-coordinates,


y_2-y_1 = Difference between two y-coordinates of same x-coordinates,

Let
(3,5.25)=(x_1,y_1) and
(3,-8.75)=(x_2,y_2).

Upon substituting our given values in above formula, we will get:


D=√((3-3)^2+(-8.75-5.25)^2)


D=√((0)^2+(-14)^2)


D=√(0+196)


D=√(196)


D=14

Therefore, the distance between the given points is 14 units and option D is the correct choice.

User Bazdin
by
7.5k points

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