Question

For all values of $\theta$ the lines represented by the equation
$(2\cos \theta +3\sin \theta )x+(3\cos \theta -5\sin \theta )y-(5\cos \theta -2\sin \theta )=0$

• A. pass through a fixed point
• B. pass through the point $(1,1)$
• C. pass through a fixed point whose reflection in the line $x+y=\sqrt{2}$ is $(\sqrt{2}-1,\sqrt{2}-1)$
• D. pass through the origin

Answer 1

Answer: (A), (B), (C)

The correct options are
A pass through a fixed point
B pass through the point $(1,1)$
C pass through a fixed point whose reflection in the line $x+y=\sqrt{2}$ is $(\sqrt{2}-1,\sqrt{2}-1)$

The given equation can be written as $(2x+3y-5)\cos \theta +(3x-5y+2)\sin \theta =0or(2x+3y-5)+\tan \theta (3x-5y+2)=0$
This passes through the point of intersection of the lines

$2x+3y-5=0$ and $3x-5y+2=0$ for all values of $\theta$.

The coordinates of the point $P$ of intersection are (1, 1).

Let $Q(h,k)$ be the reflection of $P(1,1)$ in the line
$x+y=\sqrt{2}$ ( 1 )
Then $PQ$ is perpendicular to (1) and the mid-point of $PQ$ lies on (1).
$\therefore \frac{k-1}{h-1}=1\Rightarrow k=h$
and $\frac{h+1}{2}+\frac{k+1}{2}=\sqrt{2}\Rightarrow h=k=\sqrt{2}-1$