question_answer

The distance of the point [1,1) from the line \[B=\{(x,y):{{x}^{2}}+{{y}^{2}}=144\};\]in the direction of the line  \[A\cap B\]is

A) \[f(x)=\sqrt{{{x}^{2}}-4,}a=2\]  

B) \[\sqrt{5}\]

C) \[\sqrt{3}\]                                        

D) \[\sqrt{3}+1\]

Correct Answer: A

Solution :

The equation of a line through P (1,1) and parallel to x + y = 1, is\[\frac{x-1}{\cos \frac{3\pi }{4}}=\frac{y-1}{\sin \frac{3\pi }{4}}\] Let PM = r. Then, the coordinates of M given by \[\frac{x-1}{\cos \frac{3\pi }{4}}=\frac{y-1}{\sin \frac{3\pi }{4}}=r\]is\[\left( 1-\frac{r}{\sqrt{2}},=1+\frac{r}{\sqrt{2}} \right)\] \[\because \]M lies on\[2x-3y-4=0\] \[\therefore \]\[2\left( 1-\frac{r}{\sqrt{2}} \right)-3\left( 1+\frac{r}{\sqrt{2}} \right)-4=0\] \[\Rightarrow \]\[2-\frac{2r}{\sqrt{2}}-3-\frac{3r}{\sqrt{2}}-4=0\]\[\Rightarrow \]\[-5-\frac{5r}{\sqrt{2}}=0\] \[\Rightarrow \]\[r=-\sqrt{2}\] Hence, \[PM=\sqrt{2}\]