Question

A straight line $L$ is drawn through the point $A(2,1)$ such that its point of intersection with $x+y=9$ is at a distance of $3\sqrt{2}$ unit's from $A$. Then

• A. Inclination of $L$ is $\frac{\pi }{4}$
• B. Inclination of $L$ is $\frac{\pi }{6}$
• C. $L$ cuts the $x-$axis at $(1,0)$
• D. $L$ passes through $(5,4)$

Answer 1

Answer: (A), (C), (D)

The correct options are
A Inclination of $L$ is $\frac{\pi }{4}$
C $L$ cuts the $x-$axis at $(1,0)$
D $L$ passes through $(5,4)$

Let the angle of inclination is $\theta$

Now, $(2+3\sqrt{2}\cos \theta ,1+3\sqrt{2}\sin \theta )$ should lie on the line $x+y=9$, so
$2+3\sqrt{2}\cos \theta +1+3\sqrt{2}\sin \theta =9\Rightarrow \cos \theta +\sin \theta =\sqrt{2}\Rightarrow \frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta =1\Rightarrow \cos (\frac{\pi }{4}-\theta )=1\Rightarrow \theta =\frac{\pi }{4}$

Now, equation of line $L$, we get
$(y-1)=\tan \frac{\pi }{4}(x-2)\Rightarrow y=x-1$
$L$ cuts $x$ axis at $(1,0)$
$L$ pass through $(5,4)$


Alternate solution:
Assuming any point the line $x+y=9$ as $(h,(9-h\left)\right)$
Now, the distance between the point and $A$ is $3\sqrt{2}$, we get
$\sqrt{(h-2{)}^2+(9-h-1{)}^2}=3\sqrt{2}\Rightarrow 2h^2-20h+68=18\Rightarrow h^2-10h+25=0\Rightarrow (h-5{)}^2=0\Rightarrow h=5$
Point of intersection of the two lines is $(5,4)$
Now, equation of the line is
$y-1=\frac{4-1}{5-3}(x-2)\Rightarrow y=x-1$