The angle between the line joining the points $(1, - 2), (3, 2)$ and the line $x + 2 y - 7 = 0$ is

π

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\frac{π}{2}

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\frac{π}{3}

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\frac{π}{6}

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Solution

The correct option is B $\frac{π}{2}$
The slope of the line joining two points $(1, - 2), (3, 2)$ is
$m_{1} = \frac{2 - (- 2)}{3 - 1} = 2$
and the slope of the given line $x + 2 y - 7 = 0$ is $m_{2} = \frac{- 1}{2}$
angle between two line is $tan θ = \frac{m_{2} - m_{1}}{1 + m_{1} m_{2}} = \frac{2 - (- \frac{1}{2})}{1 + 2 \times (- \frac{1}{2})} = \infty$
so $θ = \frac{π}{2}$

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\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1}

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The straight lines joining the origin to the points of intersection of the line 2x + y = 1 and curve $3 x^{2} + 4 x y - 4 x + 1 = 0$ include an angle

(2, \frac{π}{3})

(1, \frac{π}{6})

f (x) = lim n \to \infty {\frac{x}{1 + {3 {tan}^{2} x}^{n}}}

f (x)

π

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} 0, & - \frac{π}{2} < x < - \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = - \frac{π}{6} \end{matrix} \begin{matrix} x, & - \frac{π}{6} < x < \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = \frac{π}{6} \end{matrix} \begin{matrix} 0, & \frac{π}{6} < x < \frac{π}{2} \end{matrix} \end{matrix}

π

(1, \frac{π}{2})

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