Question

the accute angle between the lines y=2x and y=-2z is​

Answer 1

Answer:

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Answer 2

y = 2x => 2x - y = 0 ... (i)

y = -2x => 2x + y = 0 ... (ii)

$m_{1} = \frac{ - b_{1} }{a _{1}} = \frac{ - 2}{ - 1} = 2 \\ m_{2} = \frac{ - b_{2} }{ a_{2} } = - 2$

$let \: \alpha \: be \: the \: acute \: angle \: between \: the \\ lines$

so, we know that

$tan \alpha = | \frac{ m_{1} - m_{2} }{1 + m_{1} m_{2} } | \\ tan \alpha = | \frac{2 - ( - 2)}{1 + (2)( - 2)} | = | \frac{2 + 2}{1 - 4} | \\ tan \alpha = | \frac{4}{ - 3} | = | \frac{ - 4}{3} | = \frac{4}{3} \\ tan \alpha = \frac{4}{3} = \frac{perpendicular}{base} \\ \\ i.e. \: for \: some \: natural \: number \: m \\ \\ perpendicular = 4m \\ base = 3m \\ \\ by \: applying \: the \: pythorean \: triplet \\ we \: get \: \\ \\ hypotenuse = 5m \\ \\ hence \: the \: acute \: angles \: of \: the \: right \\ \: triangle \: are \: {37}^{o} \: and \: {53}^{o} \\ \\ since \: \alpha \: is \: the \: angle \: opposite \: to \: side \\ \: of \: measure \: 4m \\ \alpha \: is \: the \: greater \: angle \: among \: the \: \\ acute \: angles \: of \: the \: right \: triangle \\ \\ (since \: angle \: opposite \: to \: greater \: side \\ \: in \: a \: triangle \: is \: larger) \\ \\ hence \\ \alpha = {53}^{o}$

Hence, the acute angle between y = 2x and y = -2x is 53°