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if theta is angle between the lines a1x+b1y+c=0,a1x+b2y+c2=0 then prove that cos theta =

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Answered by deshrajsingh1982

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Answered by rinayjainsl

Answer:

It is proved that

$\cos( \theta) = | \frac{ a_{1} a_{2} + b_{1} b_{2} }{ \sqrt{a_{1} {}^{2} + b_{1} {}^{2} }. \sqrt{a_{2} {}^{2} + b_{2} {}^{2} } } |$

in the steps given below

Step-by-step explanation:

The correct question is

"if is angle between the lines a1x+b1y+c1=0,a1x+b2y+c2=0 then prove that

$\cos( \theta) = | \frac{ a_{1} a_{2} + b_{1} b_{2} }{ \sqrt{a_{1} {}^{2} + b_{1} {}^{2} }. \sqrt{a_{2} {}^{2} + b_{2} {}^{2} } } |$

The given lines are

$a_{1}x+b_{1}y+c_{1}=0 \\ a_{2}x+b_{2}y+c_{2}=0$

For a line ax+by+c=0,The slope is given by the relation

$m = \tan(\theta) = \frac{ - a}{b}$

Hence the slopes for above lines are written as follows

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

Now the angle between the given lines is

$\theta = | θ _{1} - θ _{2} |$

Applying tan function on both sides we get

$tan θ =tan| θ _{1} -θ _{2}|\\=|\frac{tan θ _{1}-tan θ _{2}}{1+ tanθ _{1} tanθ _{2}}|$

Writing the above expression in terms of slopes of lines we get the following

$tan\theta=|\frac{-\frac{a_{1}}{b_{1}} - \frac{-a_{2}}{b_{2}}}{1+(\frac{-a_{1}}{b_{1}})(\frac{-a_{2}}{b_{2}})}| \\ = > tan\theta = | \frac{ - a_{1}b_{2} + a_{2}b_{1}}{a_{1}a_{2} + b_{1}b_{2}} | \\ = > tan\theta=| \frac{ a_{2}b_{1} - a_{1}b_{2}}{a_{1}a_{2} + b_{1}b_{2}} |$

Since the value of tanθ is known,we can find the value of cosθ.From the above expression,the opposite side of θ is

$| a_{2}b_{1} - a_{1}b_{2} |$

And the adjacent side is

$|a_{1}a_{2} + b_{1}b_{2}|$

Hence the hypotenuse for the angle is

$\sqrt{| a_{2}b_{1} - a_{1}b_{2} |^{2}+|a_{1}a_{2} + b_{1}b_{2}| ^{2}} \\ =[tex]\sqrt{(a_{1}a_{2})^{2}+(b_{1}b_{2})^{2}+(a_{2}b_{1})^{2}+(a_{1}b_{2})^{2}} \\ = \sqrt{(a_{1}^{2} + b_{1}^{2}).(a_{2}^{2}+b_{2}^{2})}$

Hence the value of cosθ becomes

$cosθ = \frac{adj}{hyp} = | \frac{ a_{1} a_{2} + b_{1} b_{2} }{ \sqrt{a_{1} {}^{2} + b_{1} {}^{2} }. \sqrt{a_{2} {}^{2} + b_{2} {}^{2} } } |$

Hence,it is proved that if angle between two lines is θ then

$\cos( \theta) = | \frac{ a_{1} a_{2} + b_{1} b_{2} }{ \sqrt{a_{1} {}^{2} + b_{1} {}^{2} }. \sqrt{a_{2} {}^{2} + b_{2} {}^{2} } } |$

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if theta is angle between the lines a1x+b1y+c=0,a1x+b2y+c2=0 then prove that cos theta =​

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if theta is angle between the lines a1x+b1y+c=0,a1x+b2y+c2=0 then prove that cos theta =