Question

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

• 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)

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 $\theta$ = 0
or (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 value 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)
$\therefore \frac{k-1}{h-1}=1\Rightarrow k=hand\frac{h+1}{2}+\frac{k+1}{2}=\sqrt{2}\Rightarrow h=k=\sqrt{2}-1$