For given point A(2, 4) and B(-3, b),if AB = | vedclass

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(A) Given the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Here,$A = (2, 4)$,$B = (

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$5$ or $3$

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(A) Given the distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.Here,$A = (2, 4)$,$B = (-3, b)$,and $AB = \sqrt{26}$.Squaring both sides,we get $AB^2 = 26$.Substituting the coordinates: $(2 - (-3))^2 + (4 - b)^2 = 26$.$(2 + 3)^2 + (4 - b)^2 = 26$.$5^2 + (16 - 8b + b^2) = 26$.$25 + 16 - 8b + b^2 = 26$.$b^2 - 8b + 41 = 26$.$b^2 - 8b + 15 = 0$.Factoring the quadratic equation: $(b - 5)(b - 3) = 0$.Therefore,$b = 5$ or $b = 3$.

For given point $A(2, 4)$ and $B(-3, b)$,if $AB = \sqrt{26}$,find the possible values of $b$.

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