Consider the equation different values of x we shall find the values of y and the tabulate t the values in the following table
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2 Consider the equation y = 2 x + 3 for different values of x we shall find the values of y and the tabulate t the values in the following table x y
3 When the points are plotted on the graph, we find that it forms the pattern of a straight line, the set of points lies on the straight line y = 2x + 3. Therefore the path traced out by the different points is a straight line, the path traced out is called the locus.
4 Locus of a point is the path traced out by a point, when it moves under certain mathematical condition / conditions. Nt Note: Locus is a Greek word. Loci is the plural form of locus.
5 1.The locus of a point moving at constant distance from a fixed point is called the circle. 2.Locus of apoint equidistant t from the line 3.If joining end points is the perpendicular bisector of the line. A and B arefixedpoints, P moves so that angle APB is a right angle, is always a circle with AB as diameter.
6 The equation of the locus is the relationship between x and y, satisfying the given mathematical condition, thus the equation of the locus is afunction of x and y.
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8 Take P(x 1,y 1 )asanypointonthelocus. 1 1 Write the geometrical condition / conditions satisfied by the point P. Express the above condition in terms of x 1 and y 1 by using appropriate formula / formulas. Simplify the relation if necessary then replace x 1 by x and y 1 by y.
9 1)Does the point (4,5) lie on the locus whose equation is x + y 6x 8y + 23 = 0. 2)Does the origin lie on the locus 3 x + y 5 x+ 7 y+ 9= 0.
10 Solution: 1)Put x=4 and y=5. We have =0. Therefore (4,5) lies on the locus. 2) The origin means the point (0,0) put x=0, y=0. We have 3(0)+0-5(0)+7(0)+9 0 = 9 0. Hence origin does not lie on the locus.
11 3) If the distance of a point from (-5,12) is 13,, prove that its locus is x + y + 10x 24y = 0.
12 Solution: Let P(x,y ) be any point on the locus. Given 1 1 that distance of P(x 1,y 1 ) from (-5,12) =13. By distance formula ( x = ) + ( y1 12) on squaring (x 1 +5) 2 +(y 1-12) 2 =169.
13 On simplifying. x + y + 10x 24y = Therefore x + y + 10x 24 y = 0. Replacing x 1 to x, y 1 to y We have, x y x y = 0
14 4) Distance from (0,-3) is twice the distance from (0,3).
15 Solution: Let P(x 1,y 1 ) be any point on the locus. Distance of P(x 1,y 1 ) from (0,-3) = 2 X Distance of P(x 1,y 1 ) from (0,3). ( x1 0) ( y1 3) + + ( x1 0) + ( y1 3) On Squaring, = 2 x y y x y y = 4( )
16 x + 3y 30y + 27= 0 2 y 2 y x + y 10y + 9 = 0 x + y 10 y+ 9 = 0
17 5) The square of its distance from (3,0)=7. Solution: Let P(x 1,y 1 ) be any point on the locus. PA 2 =7, ( x 3) 2 + ( y 0) 2 = x + y 6 x + 2 = 0 x + y 6 x + 2 = 0
18 6) Two vertices of a triangle are (3,4) and (5,7) and the area of the triangle is 9 sq.units. Find the locus of the third vertex.
19 Solution: Let P(x 1,y 1 ) be any point on the locus. Given that PAB forms a triangle of area 9sq.units i.e. 1 [ x 1 ( 4 7 ) + 3 (7 y 1 ) + 5 ( y 1 4 )] = x y y = x y = 0 3 x 1 2 y = 0 3 x 2 y = 0
20 7) Find the locus of the point which always moves on the line joining points (2,-3) and (3,2)
21 Solution: Let P(x 1,y 1 ) be any point on the locus. Given that P(x 1,y 1 ) lies on the line joining points (2,-3) and (3,2). Therefore the points are collinear. Hence 1 [ x1 ( 3 2) + 2(2 y1 ) + 3( y1 + 3)] = 0 2 5x y1 + 3y1 + 9 = 0 5x1 + y = 0 5x1 y1 13 = 0 5 x y 13 = 0
22 8) A point moves such that the difference of the squares of its distance from the points (4,5) and (-1,2) is 7. Find the equation of the locus
23 Solution: Let P(x 1,y 1 ) be any point on the locus. Given that PA PB = 7 ( x y ) ( x y ) ( 4) + ( 5) ( + 1) + ( 2) = 7 x1 8x y1 10y ( x1 + 2x y1 4y ) = 0 10x1 6y = 0 10x+ 6y 29 = 0
24 9) The co ordinates of two fixed points A and B are (a,0) and (-a,0) respectively. Find the equation of the locus of the point P when AB subtends right angle at P.
25 Solution: Let P(x 1,y 1 ) be any point on the locus. Given that AB subtends right angle at P i.e APB is a right angled triangle. 2 AB = AP + PB ( ) ( a + a) = 2 x + 2 y + 2a x + y = a x + y = a
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