Question

find the angle between the lines 3x +y-7=0 and x+2y+9=0

Answer 1

first find slopes of both lines
slope of 3x+y-7=0 is m1=-3
and
slope of x+2y+9=0 is m2=-1/2

now use formula
tan@=|m1-m2|/|1+m1m2|
=(-1/2+3)/|(1-3/2)|=5
@=tan^-1 (5)

Answer 2

Answer:

$\theta = tan^{-1}(\frac{-7}{5})$

Step-by-step explanation:

General equation of line is ax + by + c = 0

So, Slope = $m = \frac{-a}{b}$

Slope of $3x + y - 7 = 0$ is $m_1 = -3$

Slope of $x + 2y + 9 = 0$ is $m_2 = \frac{-1}{2}$

To find the angle between two lines

$tan\theta =| \frac{(m_2 - m_1)}{(1 + m_1m_2)}|$

$tan \theta = | \frac{\frac{-1}{2}-3}{ (1+\frac{-1}{2} \times (-3))}|$

$tan \theta = | \frac{\frac{-7}{2}}{1+\frac{3}{2}} |$

$tan \theta = | \frac{\frac{-7}{2}}{\frac{5}{2}} |$

$tan \theta = | \frac{-7}{5} |$

$\theta = tan^{-1}(\frac{-7}{5})$

Hence the angle is $\theta = tan^{-1}(\frac{-7}{5})$

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