Question

$A(3,0)$ and $B(6,0)$ are two fixed points and $U(x_1,y_1)$ is a variable point of the plane. $AU$ and $BU$ meet the $y-$axis at $C$ and $D$, respectively, and $AD$ meets $OU$ at $V$. Then for any position of $U$ in the plane $CV$ passes through fixed point $(p,q)$, whose distance from origin is (where $O$ is origin)

• A. $1unit$
• B. $2unit$
• C. $3unit$
• D. $4unit$

Answer 1

The correct option is B $2unit$

Given points are,

$A(3,0)andB(6,0)$are two fixed points $U(x_1,y_1)$.

Equation of line $AU=y-y_1=\frac{0-y_1}{3-x_1}(x-x_1)$

This cut $yaxis$ so, $x=0$

$y=\frac{3y_1}{3-x_1}$

So, coordiantes C is $(0,\frac{3y_1}{3-x_1})$

Equation of line $\begin{align}

$BU\Rightarrow y-y_1=\frac{0-y_1}{6-x_1}(x-x_1)$

This cut$y-axis$ So, $x=0$

Coordinates of point $D(0,\frac{6y_1}{6-x_1})$

Now,

Equation of line AD$\Rightarrow$ $\frac{x}{3}+\frac{y}{6y_1}(6-x_1)=1......\left(1\right)$

Equation of line $AU$ passes through the points $(0,0)and(x_1,y_1)$

So,

Equation of line

$OU\Rightarrow y=\frac{y_1}{x_1}x$

$\Rightarrow x{y}_1=y{x}_1......\left(2\right)$

Now,

$\frac{x_{1}y}{3y_1}+\frac{y(6-x_1)}{6y_1}=1$

$\Rightarrow \frac{x_{1}y}{3y_1}+\frac{6y}{6y_1}-\frac{y{x}_1}{6y_1}=1$

So,

$y=\frac{6x_1}{6+x_1},x=\frac{6x_1}{6+x_1}$

Now, Point $V\text{is}(\frac{6x_1}{6+x_1},\frac{6y_1}{6+x_1})$

So, equation of $CVis$

$y-\frac{3y_1}{3-x_1}=\frac{\frac{6y_1}{6+x_1}-\frac{3y_1}{3-x_1}}{\frac{6x_1}{6+x_1}-0}(x-0)$

$\Rightarrow y-\frac{3y_1}{3-x_1}=\frac{6y_1(3-x_1)-3y_1(6+x_1)}{(3-x_1)6x_1}.x$

$\Rightarrow y=\frac{3y_1}{3-x_1}-\frac{9x_1{y}_1}{6x_1(3-x_1)}.x$

$\Rightarrow y=\frac{3y_1}{3-x_1}[1-\frac{1}{2}x]$

Put $y=0,\text{so}1-\frac{x}{2}=0$

$x=2$

Hence, this is the answer.