The point on the X-axis which is equidistant | vedclass

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(C) Let the point on the $X$-axis be $P\ (x, 0, 0)$.Since $P$ is equidistant from $A\ (2, -5, 7)$ and $B\ (1, 3, 6)$,we have $PA = PB$,which implies $

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$P(16, 0, 0)$

Vedclass · Answer

(C) Let the point on the $X$-axis be $P\ (x, 0, 0)$.Since $P$ is equidistant from $A\ (2, -5, 7)$ and $B\ (1, 3, 6)$,we have $PA = PB$,which implies $PA^2 = PB^2$.Using the distance formula: $(x - 2)^2 + (0 - (-5))^2 + (0 - 7)^2 = (x - 1)^2 + (0 - 3)^2 + (0 - 6)^2$.$(x - 2)^2 + 25 + 49 = (x - 1)^2 + 9 + 36$.$x^2 - 4x + 4 + 74 = x^2 - 2x + 1 + 45$.$x^2 - 4x + 78 = x^2 - 2x + 46$.$-4x + 2x = 46 - 78$.$-2x = -32$.$x = 16$.Thus,the point is $P\ (16, 0, 0)$.

The point on the $X$-axis which is equidistant from $A\ (2, -5, 7)$ and $B\ (1, 3, 6)$ is . . . . . . .

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