Question

A(3,0) and B(6,0) are two fixed points and U() is a variable point of the plane .AU and BU meets 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​

Answer 1

x=2 UNITS .......HOW ?SEE DOWN

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 =

3−x 1

0−y 1

(x−x 1)

This cut yaxis so, x=0

y= 3−x 1

3y 1

So, coordiantes C is (0, 3−x 1 3y 1 )

Equation of line $$\begin{align}

BU⇒y−y 1

=

6−x 1

0−y 1

(x−x 1 )

This cuty−axis So, x=0

Coordinates of point D(0, 6−x 1 6y 1 )

Now,

Equation of line AD⇒

3 x+ 6y 1

y (6−x 1 )=1......(1)

Equation of line AU passes through the points (0,0)and(x 1 ,y 1 )

So,

Equation of line

OU⇒y= x 1 y 1

x

⇒xy 1 =yx 1......(2)

Now,

3y 1 x 1

y + 6y 1

y(6−x 1 )

=1

⇒ 3y 1

x 1 y + 6y 1

6y − 6y 1

yx 1

=1

So,

y= 6+x 1

6x 1

,x= 6+x 1

6x 1

Now, Point Vis( 6+x 1 6x 1 , 6+x 1 6y 1 )

So, equation of CVis

y− 3−x 13y 1

= 6+x 1 6x 1 −0 6+x 1

6y 1 − 3−x 1 3y 1

(x−0)

⇒y− 3−x 1 3y 1

=

(3−x 1 )6x 1

6y 1

(3−x 1 )−3y 1

(6+x 1 ) .x

⇒y= 3−x 1

3y 1− 6x 1

(3−x 1 )9x 1 y 1 .x

⇒y= 3−x 1

3y 1

[1− 2 1 x]

Put y=0,

so1− 2x =0

x=2

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