Question

distance of the point p(x,12) from the origin is 13units . if the point P is in the second quadrant then what is the value of x​

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: point \: P \: is \: at \: a \: distance \: of \: 13 \: units \: from \: the \: origin$

$so \: as \: point \: P \: lies \: in \: second \: quadrant \: x \: coordinate \: would \: be \: negative \:$

$so \: let \\ (x,12) = (x,y) \\ now \: using \: origin \: distance \: formula \\ we \: get \\ d \: (P) = \sqrt{x {}^{2} + y {}^{2} } \\ 13 = \sqrt{x {}^{2} + (12) {}^{2} } \\ squaring \: both \: sides \\ we \: get \\ 169 = x {}^{2} + 144 \\ x {}^{2} = 25 \\ ie \: x = 5 \: \: or \: \: x = - 5$

$but \: as \: point \: P \: lies \: in \: quadrant \: 2 \: \\ its \: x \: coordinate \: is \: negative \: thus \: x = 5 \: is \: absurd \\ hence \: x = - 5 \: units$