Question

Find the values of x for which the distance between the points P (2, -3) and Q (x, 5) is 10.​

Answer 1

Answer:

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Step-by-step explanation:

$PQ = \sqrt{( {x_2-x_1})^{2} + {(y_2-y_1)}^{2} } \\ 10 = \sqrt{ {(x - 2)}^{2} + {(5 + 3)}^{2} } \\ 10 = \sqrt{ {x}^{2} + 4 - 4x + 64 } \\ {10}^{2} = {x}^{2} - 4x + 68 \\ = > {x}^{2} - 4x - 32 \\ = > {x}^{2} - 8x + 4x - 32$

$= > x(x - 8) + 4(x - 8) \\ = > (x + 4)(x - 8) \\ x = - 4,8$