If P(4, -7) and Q(-1, 5),then PQ = dots | vedclass

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Question

(A) To find the distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$,we use the distance formula:$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$Giv

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) To find the distance between two points $P(x_1, y_1)$ and $Q(x_2, y_2)$,we use the distance formula:$PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$Given $P(4, -7)$ and $Q(-1, 5)$,we have $x_1 = 4, y_1 = -7, x_2 = -1, y_2 = 5$.Substituting these values into the formula:$PQ = \sqrt{(-1 - 4)^2 + (5 - (-7))^2}$$PQ = \sqrt{(-5)^2 + (5 + 7)^2}$$PQ = \sqrt{(-5)^2 + (12)^2}$$PQ = \sqrt{25 + 144}$$PQ = \sqrt{169}$$PQ = 13$Therefore,the distance $PQ$ is $13$.

If $P(4, -7)$ and $Q(-1, 5)$,then $PQ = \dots$

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