a) Find the equation of the locus of the point which moves so that it is always equidistance from the points (1, 3) and ( 2, 6).

Equation of the Locus of a Point Equidistant from Two Given Points

Step 1: Understanding the Problem

We are given two fixed points in a Cartesian plane:

• Point $A(1,3)$
• Point $B(-2,6)$

We need to find the equation of the path (locus) of a point $P(x,y)$ that moves such that its distance from $A$ is always equal to its distance from $B$.

Step 2: Using the Distance Formula

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ in a plane is given by: $$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$$

Let $P(x,y)$ be the moving point. The condition that $P$ is equidistant from $A$ and $B$ can be written as: $$PA=PB$$ $$\sqrt{(x-1{)}^2+(y-3{)}^2}=\sqrt{(x+2{)}^2+(y-6{)}^2}$$

Step 3: Squaring Both Sides to Eliminate Square Roots

To simplify, we square both sides: $$(x-1{)}^2+(y-3{)}^2=(x+2{)}^2+(y-6{)}^2$$

Step 4: Expanding the Squared Terms

We expand each squared term: $$(x^2-2x+1)+(y^2-6y+9)=(x^2+4x+4)+(y^2-12y+36)$$

Step 5: Simplifying the Equation

Combine like terms on both sides: $$x^2-2x+y^2-6y+10=x^2+4x+y^2-12y+40$$

Subtract $x^2+y^2$ from both sides: $$-2x-6y+10=4x-12y+40$$

Bring all terms to one side: $$-2x-6y+10-4x+12y-40=0$$ $$-6x+6y-30=0$$

Step 6: Simplifying Further

Divide the entire equation by $-6$: $$x-y+5=0$$

Step 7: Rewriting in Standard Form

The equation can be written as: $$y=x+5$$

Step 8: Interpretation of the Result

The equation $y=x+5$ represents a straight line. This means the locus of the point $P(x,y)$ that is always equidistant from $A(1,3)$ and $B(-2,6)$ is a straight line.

Verification

To ensure correctness, let's verify with a point on the line, say $(0,5)$:

• Distance from $A(1,3)$: $$\sqrt{(0-1{)}^2+(5-3{)}^2}=\sqrt{1+4}=\sqrt{5}$$
• Distance from $B(-2,6)$: $$\sqrt{(0+2{)}^2+(5-6{)}^2}=\sqrt{4+1}=\sqrt{5}$$ Both distances are equal, confirming the correctness of the equation.

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Final Answer

The equation of the locus of the point $P(x,y)$ that is always equidistant from $(1,3)$ and $(-2,6)$ is: $$\boxed{y=x+5}$$