Find the distance d( $$\mathrm { d } \left( \mathrm { P } _ { 1 } , \mathrm { P } _ { 2 } \right) \text { between the points } \mathrm { P } _ { 1 } \text { and } \mathrm { P } _ { 2 }$$ .
-$$\mathrm { P } _ { 1 } = ( 5 , - 7 ) ; \mathrm { P } _ { 2 } = ( 7 , - 1 )$$
A) 4
B) 32
C) $2 \sqrt { 10 }$
D) $32 \sqrt { 2 }$

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 $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Applying this formula to points $P_1 = (5, -7)$ and $P_2 = (7, -1)$, we get $\sqrt{(7 - 5)^2 + (-1 + 7)^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$.

Find the Distance D - P1=(5,−7);P2=(7,−1)\mathrm { P } _ { 1 } = ( 5 , - 7 ) ; \mathrm { P } _ { 2 } = ( 7 , - 1 )P1​=(5,−7);P2​=(7,−1)

Find the Distance D
- $\mathrm { P } _ { 1 } = ( 5 , - 7 ) ; \mathrm { P } _ { 2 } = ( 7 , - 1 )$