A locus or curve is the set of points and only those points satisfying a given condition or a well defined property. This is the locus definition in geometry. Example of locus is provided below:
(1) A,B are two fixed points. Let a set of points equidistant from A and B. Locus is the perpendicular bisector of AB . Every point on the perpendicular bisector of AB obeys the condition.
(2) The locus of a set of points which are at a constant distance from a fixed point,is a circle.
- Every point satisfying the given conditions lies on the locus.
- Every point on the locus satisfies the given conditions.
- The locus can be a straight line or a curved line (lines).
- State what is given and the condition to be satisfied.
- Find several points satisfying the condition, which indicate the shape of the locus.
- Connect the points and describe the locus fully.
The equation to a locus is the condition which the coordinates of each of the points of that locus and only those points must satisfy. In other words, the equation to a locus is nothing but the geometrical property expressed in algebraic language.
Conversely, let (x,y) represent any point on the locus, x and y satisfy the equation of the locus and every point satisfying the equation lies on the locus.
If the functional relation between x,y be f (x,y) = 0, then we say f (x,y ) = 0 is the Cartesian equation of the locus.
Thus in coordinate geometry a locus is represented by an equation. Let us see locus of equation in locus learning.
The set of points (x,y) ,satisfying a given equation f (x,y) = 0 is called the locus or graph of the equation.
In other words, if f is a relation on R, then the graph of f is the set of all (x,y) `in` R, in a coordinate plane.
e.g., The set of a points which are at a constant distance of 5 units from a fixed point (2,3) has the equation
sqrt(( x -2 )2 + ( y - 3 )2) = 5
=> x2 + y2 - 4x - 6y - 12 = 0
Below are some solved problems on locus
Q 1: Find the locus of the point which is at a constant distance of 10 units from (2,-8) .
solution : P(x,y) `in` Locus. If A = (2,-8) , given geometrical condition is PA = 10
=> PA2 = 100
=> (x -2 2) + ( y+8 2) = 100
=>Equation of locus is x2 + y2 - 4x +16y - 32 = 0
Q 2: Find the equation to the set of points equidistant from (-1,-1)and (4,2).
solution : Let A =( -1,-1) and B =( 4,2)
If P( x,y ) `in ` set then the geometrical condition satisfied by P is
PA = PB
=> PA2 = PB2
=> ( x +1) 2 + ( y + 1 )2 = ( x - 4 )2 + (y - 2) 2
=> x2 +2x +1 +y2 +2y +1 = x2 - 8x + 16 + y2 - 4y + 4
=> 5x + 3y = 9
Therefore, the equation of locus is 5x + 3y = 9
Q 3: Find the locus of a point which moves in such a way that its distance from (3,0) is always equal to its distance from the origin.
solution : Let A = (3,0) and O = (0,0).
Let P( x,y ) be a point satisfying the geometrical condition.,
PA = OP
=> PA2 = OP2
=> ( x - 3 )2 + y2 = x2 +y2
=> 6x = 9
Therefore, the equation of locus of P is 2x - 3 = 0.
Q 4: If p is the length of perpendicular drawn from the origin on the line AB whose intercepts on the axes are a and b, then show
Equation of the line AB in the intercept form is
Since (i) and (ii) represent the same line, we have
Squaring and adding, we get
Q 5: A straight line passes through the point (a, b) and this bisects the part of the line intercepted between the axes.
Let the straight line passing through p(a, b) cut the x-axis at A(a,0) and B(0,b) on the x-axis and the y-axis respectively.
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