Find the values of y for which the distance b | vedclass

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(A) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Given points are $P(2

Vedclass · Accepted Answer

$y = 3, -9$

Vedclass · Answer

(A) The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Given points are $P(2, -3)$ and $Q(10, y)$,and the distance $d = 10$.Substituting the values into the formula:$10 = \sqrt{(10 - 2)^2 + (y - (-3))^2}$$10 = \sqrt{(8)^2 + (y + 3)^2}$Squaring both sides:$100 = 64 + (y + 3)^2$$100 - 64 = (y + 3)^2$$36 = (y + 3)^2$Taking the square root on both sides:$y + 3 = \pm 6$Case $1$: $y + 3 = 6 \implies y = 3$Case $2$: $y + 3 = -6 \implies y = -9$Therefore,the possible values of $y$ are $3$ and $-9$.

Find the values of $y$ for which the distance between the points $P(2, -3)$ and $Q(10, y)$ is $10$ units.

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