Find the distance d( $$\mathrm { d } \left( \mathrm { P } _ { 1 } , \mathrm { P } _ { 2 } \right) \text { between the points } \mathrm { P } _ { 1 } \text { and } \mathrm { P } _ { 2 }$$ .
-$\mathrm { P } _ { 1 } = ( 0.3 , - 0.2 ) ; \mathrm { P } _ { 2 } = ( 2.1 , - 1.3 )$ Round to three decimal places, if necessary.
A) $2.21$
B) $2.11$
C) $6.671$
D) $14.5$

Question

Quizplus · Accepted Answer

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a plane is given by the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Applying this formula to the given points $P_1 = (0.3, -0.2)$ and $P_2 = (2.1, -1.3)$, we get $d = \sqrt{(2.1 - 0.3)^2 + (-1.3 + 0.2)^2} = \sqrt{(1.8)^2 + (-1.1)^2} = \sqrt{3.24 + 1.21} = \sqrt{4.45} \approx 2.11$.

Find the Distance D - P1=(0.3,−0.2);P2=(2.1,−1.3)\mathrm { P } _ { 1 } = ( 0.3 , - 0.2 ) ; \mathrm { P } _ { 2 } = ( 2.1 , - 1.3 )P1​=(0.3,−0.2);P2​=(2.1,−1.3)

Find the Distance D
- $\mathrm { P } _ { 1 } = ( 0.3 , - 0.2 ) ; \mathrm { P } _ { 2 } = ( 2.1 , - 1.3 )$