Let coordinates of the points A and B are (5 , 0) and (0 , 7) respectively. P and Q are the variable points lying on the x-and y-axis respectively so that PQ is always perpendicular to the line AB. Then locus of the point of intersection of BP and AQ is         

Let the equation of PQ be $\frac{x}{a}+\frac{y}{b}=1$ ---------- (i)                                                                          

  Then equation of BP is $\frac{x}{a}+\frac{y}{7}=1$        ----------- (ii)                                                                        

  and equation of AQ is $\frac{x}{5}+\frac{y}{b}=1$          --------- (iii)                                                                            

   Since $AB\perp PQ$ , therefore $-\frac{b}{a}\times -\frac{7}{5}=-1$                                                                                        

    $\Rightarrow 5a+7b=0$                                                                                                                  

  Now, slope of BP is $-\frac{7}{a}$  and slope of AQ is $-\frac{b}{5}$                                                                              

  So, product of these slopes is $\frac{7b}{5a}$  = -1   [from (iv)]                                                                 

 $\therefore$  BP and AQ intersect each other at right angle. So, the locus of 

the point of intersection of BP and AQ is a circle with AB as diameter. Thus equation of locus is                                                                  $x(x-5)+y(y-7)=0\Rightarrow x^2+y^2-5x-7y=0$