Q. 465.0( 1 Vote )

Fill in the blank

Answer :

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

Let (p, q) be any moving point.


Distance between (p, q) and (3, -2)



(d1)2 = (p – 3)2 + (q + 2)2


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







According to the question, we have (d1)2 = d2



Taking numerical values only, we have



13[(p – 3)2 + (q + 2)2] = 5p – 12q – 3


13[p2 + 9 – 6p + q2 + 4 + 4q] = 5p – 12q – 3


13p2 + 117 – 78p + 13q2 + 52 + 52q = 5p – 12q – 3


13p2 + 13q2 – 78p + 52q + 169 = 5p – 12q – 3


13p2 + 13q2 – 78p + 52q + 169 – 5p + 12q + 3 = 0


13p2 + 13q2 – 83p + 64q + 172 = 0


Ans. 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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