A point moves so that it is always equidistant from (-5, 5) and (-2, 2). Find the equation of its locus.
Answers
Answer:
What is the equation of the its locus when a point moves so that it is always equidistant from (-5,5) and (-2,2)?
The locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining these two fixed points.
In this case, the two fixed points are given as A(−5,5) and B(−2,2).
⇒ The midpoint of AB is (−72,72).
The slope of AB is (5−2)(−5+2)=−1.
⇒ The slope of the line perpendicular to AB is 1.
⇒ The perpendicular bisector of AB passes through point (−72,72) and has a slope 1.
⇒ The equation of the perpendicular bisector of AB is y−72=1(x+72).
⇒ The equation of the perpendicular bisector of AB is x−y+7=0.
Step-by-step explanation:
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Given : A point moves so that it is always equidistant from (-5, 5) and (-2, 2).
To Find : the equation of its locus.
Solution:
Let say Point P (x , y) is equidistant from A (-5, 5) and B (-2, 2).
PA = PB
PA² = PB²
=> ( x- (-5))² + ( y - 5)² = ( x - (-2))² + ( y - 2)²
=> x² + 10x + 25 + y² -10y + 25 = x² + 4x + 4 + y² - 4y + 4
=> 6x - 6y + 42 = 0
=> x - y + 7 = 0
x - y + 7 = 0 is the locus of point equidistant from (-5, 5) and (-2, 2).
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