The angles between the lines $3 x + y - 7 = 0 and x + 2 y + 9 = 0$ is:

\frac{π}{3}

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\frac{π}{6}

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\frac{π}{2}

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\frac{π}{4}

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Solution

The correct option is B $\frac{π}{4}$
Slope of the line $y = - 3 x + 7$ is $- 3$
Slope of the line $2 y = - x - 9$ or $y = \frac{- 1}{2} x - \frac{9}{2}$ is $- \frac{1}{2}$

$∴ m_{1} = - 3, m_{2} = - \frac{1}{2}$

Angle between the line $= tan θ = ∣ ∣ ∣ \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ ∣ ∣ \frac{- 3 + \frac{1}{2}}{1 + (- 3) (- \frac{1}{2})} ∣ ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ ∣ \frac{\frac{- 5}{2}}{1 + \frac{3}{2}} ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ ∣ \frac{\frac{- 5}{2}}{\frac{2 + 3}{2}} ∣ ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ ∣ \frac{\frac{- 5}{2}}{\frac{5}{2}} ∣ ∣ ∣ ∣ ∣$

$tan θ = 1$

$∴ θ = \frac{π}{4}$

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Similar questions

Q. The angle between the lines

3 x + y - 7 = 0

and

x + 2 y + 9 = 0

Q. Find the angle between the lines

3 x + y - 7 = 0

and

x + 2 y - 9 = 0

Q. Find the acute angle between the lines

3 x + y - 7 = 0

and

x + 2 y - 9 = 0.

$π_{1}, π_{2}, π_{3} a n d π_{4}$ are the osmotic pressures of 5% $(\frac{W}{V})$ solutions of urea, fructose, sucrose and KCl respectively at certain temperatures. The correct order of their magnitude is

\int_{0}^{3} \frac{3 x + 1}{x^{2} + 9} d x =

(a)

\frac{π}{12} + \log (2 \sqrt{2})

(b)

\frac{π}{2} + \log (2 \sqrt{2})

(c)

\frac{π}{6} + \log (2 \sqrt{2})

(d)

\frac{π}{3} + \log (2 \sqrt{2})

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The angles between the lines 3x+y−7=0 and x+2y+9=0 is:

The angles between the lines $3 x + y - 7 = 0 and x + 2 y + 9 = 0$ is: