Question

The distance between (p, - q) and (q, - p) is (p > q) .​

Answer 1

Answer:

The distance between (p, - q) and (q, - p) is $\sqrt{2q^2+2p^2-4pq}$

Step-by-step explanation:

If P$(x_1, y_1)$ and Q$(x_2, y_2)$ are the two points in a plane, then the distance between P and Q can be given by

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

SO the distance between $(p, - q) and (q, - p)$ is

distance=

                          $\sqrt{(q-p)^{2} +(p-q)^{2} }\\ \\ \sqrt{q^2+p^2-2pq+p^2+q^2-2pq} \\ \\ \sqrt{2q^2+2p^2-4pq}$

                                 

Answer 2

Answer: orStep-by-step explanation:if the answer is correct please mark me as a brainlist