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If a variable line drawn through the intersection of the lines \frac{x}{3} + \frac{y}{4} = 1\; and\: \frac{x}{4} + \frac{y}{3} = 1,

 meets the coordinate axes at A and B, (A ≠ B), then the locus of the midpoint of AB is :

  • Option 1)

    6xy = 7(x + y)

  • Option 2)

    4( x + y )2 − 28( x + y ) + 49=0

  • Option 3)

    7xy = 6( x + y )

  • Option 4)

    14( x + y )2 − 97( x + y ) + 168 = 0

 

Answers (1)

best_answer

As we have learned

Section formula -

x= \frac{mx_{2}+nx_{1}}{m+n}

y= \frac{my_{2}+ny_{1}}{m+n}

- wherein

If P(x,y) divides the line joining A(x1,y1) and B(x2,y2) in ration m:n

 

 

Family of straight lines -

L_{1}+\lambda L_{2}=0


 

- wherein

 L_{1}\, and \, L_{2}=0 are the equations of the lines and \lambda is a constant.

 

 

Family of lines 

(4x+3y-12)+\lambda (3x+4y-12) = 0

Now for x = 0  (3+4 \lambda ) y = 12 (1+ \lambda )

\Rightarrow y = \frac{12 (1+ \lambda )}{3+4\lambda }

For y = 0 (4+3 \lambda )x = 12 (1+ \lambda ) \\x = \frac{12 (1+ \lambda )}{4+3\lambda }

 

h = \frac{6 (1+ \lambda )}{4+ 3 \lambda } \: \: \: and \: \: \: \\ K = \frac{6 (1+\lambda )}{3+4 \lambda }

This (h,k ) satisfies 

 A_{xy} = 6 (x+y )

 

 

 

 

 


Option 1)

6xy = 7(x + y)

Option 2)

4( x + y )2 − 28( x + y ) + 49=0

Option 3)

7xy = 6( x + y )

Option 4)

14( x + y )2 − 97( x + y ) + 168 = 0

Posted by

Himanshu

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