A) \[{{90}^{o}}\]
B) \[{{45}^{o}}\]
C) \[{{180}^{o}}\]
D) \[{{60}^{o}}\]
Correct Answer: A
Solution :
Lines are \[p=\left| \frac{-k}{\sqrt{{{\sec }^{2}}\alpha +\text{cose}{{\text{c}}^{2}}\alpha }} \right|\] .....(i) and \[3x+y+4=0\] ......(ii) Here, \[{{m}_{1}}=2,\,{{m}_{2}}=-\,3\] If angle between them is \[\theta \], then \[\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\] \[=\left| \frac{2+3}{1-6} \right|=\left| \frac{5}{-5} \right|\]= 1 \[\tan \theta =\tan \frac{\pi }{4}\] Þ \[\theta =\frac{\pi }{4}=45{}^\circ .\]
You need to login to perform this action.
You will be redirected in
3 sec
