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  • All lines represented by the equation <img alt="\mathrm{(2 \cos \theta+3 \sin \theta) x+(3 \cos \theta-5 \sin \theta) y=5 \cos \theta-2 \sin \theta }" src="https://entrancecorner.onc

All lines represented by the equation

\mathrm{(2 \cos \theta+3 \sin \theta) x+(3 \cos \theta-5 \sin \theta) y=5 \cos \theta-2 \sin \theta }
pass through a fixed point for all \mathrm{\theta}. What are the coordinates of this fixed point and its reflection in the line \mathrm{x+y=\sqrt{2},} respectively?

Option: 1

(1,1)\text{ and }(\sqrt{2}, \sqrt{2})


Option: 2

(1,1) \text{ and } (\sqrt{2}+1, \sqrt{2}+1)


Option: 3

(1,1) \text{ and } (\sqrt{2}-1, \sqrt{2}-1)


Option: 4

(\sqrt{2}, \sqrt{2}) \text{ and } (1,1)


Answers (1)

best_answer

The equation (1) can be expressed as

\mathrm{(2 x+3 y-5) \cos \theta+(3 x-5 y+2) \sin \theta=0 }

or \mathrm{ \quad(2 x+3 y-5)+\tan \theta(3 x-5 y+2)=0}

It is clear that these lines will pass through the point of intersection of the lines

\begin{aligned} & \mathrm{2 x+3 y-5=0 }\\ & \mathrm{3 x-5 y+2=0 }\end{aligned}

For all values of \theta. Solving equations (3) and (4) we get the coordinates of the required fixed points as P(1,1).

Let \mathrm{Q(\alpha, \beta)} be the reflection of P(1,1) in the line \mathrm{x+y=\sqrt{2}}. The line P Q is perpendicular to the line \mathrm{x+y=\sqrt{2}} and the mid point of segment P Q lies on the line.

 \mathrm{x+y=\sqrt{2}}. Thus

\begin{array}{lll} \mathrm{\left(\frac{\beta-1}{\alpha-1}\right)(-1)=-1} & \Rightarrow & \\ \mathrm{\frac{\beta+1}{2}+\frac{\alpha+1}{2}=\sqrt{2}} & \Rightarrow & \beta=\alpha=\sqrt{2}-1 \end{array}
and
Hence coordinates of the reflection Q of P in the line \mathrm{x+y=\sqrt{2}} are

\mathrm{Q \equiv(\sqrt{2}-1, \sqrt{2}-1)}
 

Posted by

Gaurav

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