Question

Family of lines represented by the equation $\left(\cos \theta \right)x+(\cos \theta -\sin \theta )y-3(3\cos \theta +\sin \theta )=0$ passes through a fixed point $M$ for all real value of $\theta$. Find $M$

• A. $(6,3)$
• B. $(3,6)$
• C. $(-6,2)$
• D. $(3,-6)$

Answer 1

Answer: (A)

$(6,3)$

Let us consider the problem:

$\left(\left(\cos \theta +\sin \theta \right)x+\cos \theta -\sin \theta \right)y-3\left(3\cos \theta +\sin \theta \right)=0$

$\Rightarrow \cos \theta \left(x+y-9\right)+\sin \theta \left(x-y-3\right)=0$

$\Rightarrow$ $\left(x+y-9\right)+\tan \theta \left(x-y-3\right)=0$

$L_1+K{L}_2=0$(pass through intersection of $L_1$ and $L_2$ for all value of $K$)

$x+y-9=0$

$\Rightarrow$ $x-y-3=0$

Hence,

$x+y=9$

$x-y=3$

hence the intersection point is $(6,3)$