question_answer

If the equation of the locus of a point equidistant from the points \[({{a}_{1}}-{{b}_{1}})\] and \[({{a}_{2}}-{{b}_{2}})\] is \[({{a}_{1}}-{{a}_{2}})x+({{b}_{1}}-{{b}_{2}})\,y+c=0\], then the value of 'c' is     AIEEE  Solved  Paper-2003

A) \[\frac{1}{2}(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})\]             

B) \[a_{1}^{2}-a_{2}^{2}+b_{1}^{2}-b_{2}^{2}\]     

C) \[\frac{1}{2}\,\,(a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2})\]

D)       \[\sqrt{a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2}}\]

Correct Answer: A

Solution :

Let \[P(\alpha ,\beta )\] be the point which is equidistant to \[A({{a}_{1}},{{b}_{1}})\] and \[B({{a}_{2}},{{b}_{2}})\]. \[\therefore \]      \[{{(PA)}^{2}}={{(PB)}^{2}}\] \[\Rightarrow {{(\alpha -{{a}_{1}})}^{2}}+{{(\beta -{{b}_{1}})}^{2}}={{(\alpha ={{a}_{2}})}^{2}}+{{(\beta -{{b}_{2}})}^{2}}\] \[\Rightarrow \]                                       \[={{\alpha }^{2}}+a_{2}^{2}\,-2\alpha {{a}_{2}}\,+{{\beta }^{2}}+b_{2}^{2}-\,2\beta {{b}_{2}}\]                                                                         Thus, the equation of locus (a,p) is But the given equation is                                                        Alternate Solution Since, the points  and  satisfy the equation so that                         ... (i) and               ...(ii) On adding Eqs. (i) and (ii), we get     \[({{a}_{1}}+{{a}_{2}})\,({{a}_{1}}-{{a}_{2}})\,+({{b}_{1}}+{{b}_{2}})\,({{b}_{1}}-{{b}_{2}})+2c=0\] \[\Rightarrow \,\,\,\,\,2c=-(a_{1}^{2}-a_{2}^{2}\,+b_{1}^{2}\,-b_{2}^{2})\] \[\Rightarrow \,\,\,\,c=\frac{1}{2}(a_{2}^{2}\,+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})\]