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.

- A locus of points is the set of points, and only those points, that satisfy given conditions.

- 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).

- To determine a locus:

- 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 locus of points equidistant from the sides of a given angles is the bisector of the angle.

- The locus of points equidistant from two given intersecting lines is the bisectors of the angles formed by the lines.

- The locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them.

- Locus of a point equidistant from a given point in a plane is a circle.

- The locus of points equidistant from two given parallel lines is a line parallel to the two lines and midway between them.

- Locus of all points at a given distance from a given line is two straight lines parallel to the given line.

**Locus equation:**

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

=> PA^{2} = 100

=> (x -2^{ 2}) + ( y+8 ^{2}) = 100

=>Equation of locus is x^{2} + y^{2} - 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

=> PA^{2 }= PB^{2}

=> ( x +1) ^{2} + ( y + 1 )^{2} = ( x - 4 )^{2} + (y - 2)^{ 2}

=> x^{2} +2x +1 +y^{2} +2y +1 = x^{2} - 8x + 16 + y^{2} - 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

=> PA^{2} = OP^{2}

=> ( x - 3 )^{2 }+ y^{2} = x^{2} +y^{2}

=> 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

**Solution: ** **Aliter:**

Equation of the line AB in the intercept form is

...(i)

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.

Show that

**Solution: **

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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