Question

For all values of $\theta$, all the lines represented by the equation $(2\cos \theta +3\sin \theta )x+(3\cos \theta -5\sin \theta )y-(5\cos \theta -2\sin \theta )=0$ passes through a fixed point, then the reflection of that point with respect to the line $x+y=\sqrt{2}$ is

• A. $(\sqrt{2}+1,\sqrt{2}+1)$
• B. $(\sqrt{2}-1,\sqrt{2}-1)$
• C. $(\sqrt{3}-1,\sqrt{3}-1)$
• D. $(\sqrt{3}+1,\sqrt{3}+1)$

Answer 1

The correct option is B $(\sqrt{2}-1,\sqrt{2}-1)$

Given:

$(2\cos \theta +3\sin \theta )x+(3\cos \theta -5\sin \theta )y-(5\cos \theta -2\sin \theta )=0$

For $\theta =0^0$

$(2\cos 0+3\sin 0)x+(3\cos 0-5\sin 0)y-(5\cos 0-2\sin 0)=0$

$(2\ast (1)+3\ast (0\left)\right)x+(3\ast (1)-5\ast (0\left)\right)y-(5\ast (1)-2\ast (0\left)\right)=0$
$\Rightarrow 2x+3y-5=0$ $.....\left(1\right)$
For $\theta ={90}^0$

$(2\cos 90+3\sin 90)x+(3\cos 90-5\sin 90)y-(5\cos 90-2\sin 90)=0$

$(2\ast (0)+3\ast (1\left)\right)x+(3\ast (0)-5\ast (1\left)\right)y-(5\ast (0)-2\ast (1\left)\right)=0$

$3x-5y+2=0$ $....\left(2\right)$

Point of intersection of $\left(1\right)$ and $\left(2\right)$ is $(1,1)$
Now, we will find the image of $(1,1)$ w.r.t to line $x+y-\sqrt{2}=0$
Let $(h,k)$ be the image:
$\Rightarrow \frac{h-1}{1}=\frac{k-1}{1}=\frac{-2(2-\sqrt{2})}{2}$

$\Rightarrow h-1=k-1=-2+\sqrt{2}$

$\Rightarrow h=k=\sqrt{2}-1$
Hence, the reflection is $(\sqrt{2}-1,\sqrt{2}-1)$

Hence, option $B$.