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  • Fill in the blanks A point moves so that square of its distance from the point (3, –2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is ____.

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A point moves so that square of its distance from the point (3, –2) is numerically equal to its distance from the line 5x – 12y = 3. The equation of its locus is ____.

Answers (1)

Given point is (3, -2) and equation of line is  5x-12y=3  

Let (p,q) be any moving point 

  Distance between them (p,q) and (3,-2)   d_{1}=\sqrt{\left ( p-3 \right )^{3}+\left ( q-\left ( -2 \right ) \right )^{2}}

  (d1)2=(p-3)2+(q+2)2  

 Now, distance of the point (p,q) from the given line 5x-12y-3=0

d=\frac{\left | Ax+By+C \right |}{\sqrt{A^{2}+B^{2}}} 

 d_{2}=\frac{\left | 5p-12q-3 \right |}{\sqrt{(5)^{2}+(12)^{2}}}= \frac{\left | 5p-12q-3 \right |}{\sqrt{25+144}}=\frac{\left | 5p-12q-3 \right |}{\sqrt{169}}

= =\frac{\left | 5p-12q-3 \right |}{13}

 Taking numerical values only, we have  (p-3)2+(q-2)2==\frac{5p-12q-3 }{13}

    13[(p-3)2+(q+2)2]=5p-12q-3   

13[p2+9-6p+q2+4+4q]=5p-12q-3   

 On solving we get 13p2+13q2-83p+64q+172=0  

A point moves so that square of its distance from the point (3,-2) is numerically equal to its distance from the line 5x-12y=3 .

 The equation of its locus is 13p2+13q2-83p+64q+172=0

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