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### High School Mathematics - 29.19 Locus and its Equations

 The path traced by a point moving under a given condition is called its locus. If a point moves according to some fixed rule, its co-ordinates will always satisfy some corresponding algebraic relation and the path(curve) of the moving points is called the locus of the point. The curve must contain all the points satisfying the given condition so that no print outside the curve satisfies the condition. Definition: A set of all points that satisfy a given geometric condition. Method to find the equation of the locus of a moving point: Let (x,y) be ay point on the locus. Properly conceive the given geometrical condition which the above point (x,y) is to satisfy. Express the said condition in an analytical relation in x and y and simplify. The simplified equation so obtained is the equation of the locus. Example: A point moves in a plane so as to remain akways equidistant from two fixed points. A(-4, -4) and B(2,8). Find the equation of the locus. Solution: A(-4,-4) and B(2,8) are the given points. Let P(x,y) be any point on the locus. Using given informations, we getPA = PB PA2 = PB2 (x+4)2 + (y+4)2 = (x-2)2 + (y-8)2 Simplifying the above we get x+2y-3 = 0 This is the required equation of locus. Answer: x+2y-3 Directions: Solve the following questions.
 Q 1: Find the equation of the locus of the point which moves so that its distance from the axis of y is away twice its distance from the origin.42 = 03x2+4y2 = 03x2-4 = 0 Q 2: Find the equation of the locus of all points P(x,y) such that the segment OP has slope 3.y-3x = 0x2-3y = 0y2-3x = 0 Q 3: Find the equation of the locus of all the points P(x,y) such that the segment Op is parallel to the segment joining P and the point (3,2).x2-3x-2y = 03y-2x = 03x-2y = 0 Q 4: FInd the locus of a point P which moves so that the sum of the square of its distance from A(3,0) and B(-3,0) is four times the distance between A and B.x2+y2 = 3x+y = 3y2 = 3 Q 5: Fnd the equation to the locus of the point P if the sum of its distances from (1,2) and (3,4) is 25 units.2x2-8x-12y = 02x2+22-8x-12y+5 = 02y-8x = 0 Q 6: Find the equation of the locus of all points equidistant from the point (4,2) and the x-axis.y2-20 = 0x2-2y2+20 = 0x2-8x-4y+20 = 0 Q 7: If O is the origin and Q is a variable point on y2 = 8x, find the locus of the mid-point of OQ.y2-4x = 0y = 0y-4x = 0 Q 8: A(1,2), B(5,3) are two fixed points. Find the locus of point P, so that PA:PB = 2:3.5x2+5y2+22x-12y-91 = 02y2-3y-91 = 02x2+ 2y2+5 = 0 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!

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