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  • If the origin lies in the acute angle between the lines <img alt="\mathrm{a_1 x+b_1 y+c_1=0 \: and\: a_2 x+b_2 y+c_2=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Ba_1%20x&amp

If the origin lies in the acute angle between the lines \mathrm{a_1 x+b_1 y+c_1=0 \: and\: a_2 x+b_2 y+c_2=0} then \mathrm{\left(a_1 a_2+b_1 b_2\right) c_1 c_2} is
 

Option: 1

equal to zero


Option: 2

equal to one


Option: 3

positive


Option: 4

negative


Answers (1)

best_answer

In \triangle \mathrm{OM}_1 \mathrm{M}_2, \angle \mathrm{M}_1 \mathrm{OM}_2 is obtuse, i.e., \cos \angle \mathrm{M}_1 \mathrm{OM}_2<0, so that

\mathrm{O M_1^2+O M_2^2-M_1 M_2^2<0}

(1)

Now the coordinates of \mathrm{\mathrm{M}_1 \: and \: \mathrm{M}_2} are

\mathrm{\left(-\frac{a_1 c_1}{a_1^2+b_1^2},-\frac{b_1 c_1}{a_1^2+b_1^2}\right) and \left(-\frac{a_2 c_2}{a_2^2+b_2^2},-\frac{b_2 c_2}{a_2^2+b_2^2}\right)} respectively.

Also,\mathrm{ O M_1=\left|\frac{-c_1}{\sqrt{a_1^2+b_1^2}}\right| \text { and } O M_2=\left|\frac{-c_2}{\sqrt{a_2^2+b_2^2}}\right| . }

Putting these values in (1), we get

\mathrm{ \frac{c_1^2}{a_1^2+b_1^2}+\frac{c_2^2}{a_2^2+b_2^2}-\frac{a_1^2 c_1^2}{\left(a_1^2+b_1^2\right)^2}-\frac{a_2^2 c_2^2}{\left(a_2^2+b_2^2\right)^2} }

\mathrm{ -\frac{b_1^2 c_1^2}{\left(a_1^2+b_1^2\right)^2}-\frac{b_2^2 c_2^2}{\left(a_2^2+b_2^2\right)^2}+\frac{2\left(a_1 a_2+b_1 b_2\right) c_1 c_2}{\left(a_1^2+b_1^2\right)\left(a_2^2+b_2^2\right)}<0 }

\mathrm{ \Rightarrow \left(a_1 a_2+b_1 b_2\right) c_1 c_2<0 }

Hence option 4 is correct.




$$

Posted by

manish

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