find the equestion of locus of point P,such that sum of its distances from co-ordinate axes is thrice its distace from origin.
Answers
Answer:
The locus consists of the single point (0,0).
An equation for this is:
x² + y² = 0
Step-by-step explanation:
Let P = (x,y).
The sum of the distances from the axes is
|x| + |y|
The distance from the origin is
√(x²+y²)
So we need
|x|+|y| = 3√(x²+y²)
=> x²+y²+2|xy| = 9(x²+y²)
=> 2|xy| = 8(x²+y²)
=> |xy| = 4(x²+y²)
=> |xy| - 8|xy| = 4(x²-2|xy|+y²)
=> -7 |xy| = 4 ( |x| - |y| )² ... (1)
The RHS is ≥ 0, while the LHS is ≤ 0. So the only way they can be equal is if they're both equal to 0.
|xy| = 0 => either x = 0 or y = 0
If x = 0, then |x| - |y| = 0 => y = 0.
If y = 0, then |x| - |y| = 0 => x = 0.
So the only possibility is x = y = 0.
The locus consists of the single point (0,0).
There are many possibilities for an equation for this locus. One is, as we have seen, equation (1). A simpler one is
x² + y² = 0