Question

The fix point through which the line $(2\cos \theta +3\sin \theta )x+(3\cos \theta -5\sin \theta )y-(5\cos \theta -2\sin \theta )=0$ passes for all values of $\theta$ is

• A. $(0,0)$
• B. $(1,1)$
• C. $(2,1)$
• D. None of these

Answer 1

The correct option is B $(1,1)$

Given

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

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

$⟹(2x+3y-5)+\tan \theta (3x-5y+2)=0$

This represents a family of straight lines which passes throgh a fixed point

$⟹2x+3y-5=0\cdots \left(1\right)$

$⟹3x-5y+2=0\cdots \left(2\right)$

$3\times \left(1\right)⟹6x+9y-15=0\cdots \left(3\right)$

$2\times \left(2\right)⟹6x-10y+4=0\cdots \left(4\right)$

$\left(3\right)-\left(4\right)$

$⟹6x+9y-15-(6x-10y+4)=0$

$⟹6x+9y-15-6x+10y-4=0$

$⟹19y-19=0$

$⟹y=1$

Put it in $\left(1\right)$

$⟹2x+3\left(1\right)-5=0$

$⟹2x-2=0$

$⟹x=1$

$\therefore$ the point is $(1,1)$