Question

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

Answer 1

Answer:

D. 14 units.

Step-by-step 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.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$, 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:

$D=\sqrt{(3-3)^2+(-8.75-5.25)^2}$

$D=\sqrt{(0)^2+(-14)^2}$

$D=\sqrt{0+196}$

$D=\sqrt{196}$

$D=14$

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

Answer 2

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


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








Hence, Option D is correct.