Question

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

Answer 1

Answer:

distance=√(p-q) ^2 +(q-p) ^2

Answer 2

Step-by-step explanation:

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

2q

2

+2p

2

−4pq

Step-by-step explanation:

If P(x_1, y_1)(x

1

,y

1

) and Q(x_2, y_2)(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} }PQ=

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

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

distance=

\begin{gathered}\sqrt{(q-p)^{2} +(p-q)^{2} }\\ \\ \sqrt{q^2+p^2-2pq+p^2+q^2-2pq} \\ \\ \sqrt{2q^2+2p^2-4pq}\end{gathered}

(q−p)

2

+(p−q)

2