# Finding Equation of Locus by Locus Definition

## A Locus and its Equation

A locus is a collection of points with the same relations. The illustration shows a locus of points whose distance from the $y$-axis is equal to its distance from the $x$-axis.

### Example 1

A point $P(x,y)$ moves so that its distance from the $y$-axis is always equal to its distance from the $x$-axis. Find the equation of the locus of $P$.

\require{AMSsymbols} \begin{align} PA &= PB \\ PA &= \sqrt{(x-x)^2+(y-0)^2} \\ &= \sqrt{y^2} \\ &= \left|y\right| \color{green}{\cdots(1)} \\ PB &= \sqrt{(x-0)^2+(y-y)^2} \\ &= \sqrt{x^2} \\ &= \left|x\right| \color{green}{\cdots(2)} \\ \left|y\right| &= \left|x\right| &\color{green}{(1)=(2)} \\ \therefore y &= \pm x \end{align}

### Example 2

A point $P(x,y)$ moves so that its distance from the $y$-axis is always double its distance from the $x$-axis. Find the equation of the locus of $P$.

\require{AMSsymbols} \displaystyle \begin{align} 2PA &= PB \\ PA &= \left|y\right| \\ PB &= \left|x\right| \\ 2 \left|y\right| &= \left|x\right| \\ \therefore y &= \pm \frac{x}{2} \end{align}

### Example 3

Write down the equation of the locus of a point $P(x,y)$ that the distance from the $x$-axis is the square of the distance from the $y$-axis.

\require{AMSsymbols} \displaystyle \begin{align} PA &= PB^2 \\ PA &= \left|y\right| \\ PB &= \left|x\right| \\ \left|y\right| &= \left|x\right|^2 \\ \therefore y &= \pm x^2 \end{align}

### Example 4

Find the equation of the locus of a point $P(x,y)$ that the distance from the $x$-axis equals the distance from $x = -1$.

\require{AMSsymbols} \displaystyle \begin{align} PA &= PB \\ &= \sqrt{(x-x)^2+(y-0)^2} \\ &= \sqrt{y^2} \\ &= \left|y\right| \\ PB &= \sqrt{(x+1)^2+(y-y)^2} \\ &= \sqrt{(x+1)^2} \\ &= \left|x+1\right| \\ \left|y\right| &=\left|x+1\right| \\ \therefore y &=x+1 \text{ and } y = -x-1 \end{align}

### Example 5

Write down the equation of the locus of a point $P(x,y)$ that the distance from $y=2$ equals the distance from $x = -1$.

\require{AMSsymbols} \displaystyle \begin{align} PA &= PB \\ PA &= \sqrt{(x-x)^2+(y-2)^2} \\ &= \sqrt{(y-2)^2} \\ &= \left|y-2\right| \\ PB &= \sqrt{(x+1)^2+(y-y)^2} \\ &= \sqrt{(x+1)^2} \\ &= \left|x+1\right| \\ \left|y-2\right| &= \left|x+1\right| \\ y-2 &= +(x+1) \leadsto y = x+3 \\ y-2 &= -(x+1) \leadsto y = -x+1 \\ \therefore y &= x+3 \text{ and } y = -x+1 \end{align}

## Circles as Loci

A circle is a locus where a point moves equally from a certain point.

### Example 6

Find the equation of the locus of a point $P(x,y)$ that moves so it is always $2$ units from the origin.

\require{AMSsymbols} \displaystyle \begin{align} \sqrt{(x-0)^2+(y-0)^2} &= 2 \\ \sqrt{x^2+y^2} &= 2 \\ \therefore x^2+y^2 &= 4 \end{align}

### Example 7

Find the equation of the locus of a point $P(x,y)$ that moves so it is always $3$ units from $(2,0)$.

\require{AMSsymbols} \displaystyle \begin{align} \sqrt{(x-2)^2+y^2} &= 3 \\ (x-2)^2+y^2 &= 3^2 \\ \therefore (x-2)^2+y^2 &= 9 \end{align}

### Example 8

Find the equation of the locus of point $P(x,y)$ that moves so that $PA : PB = 2 : 1$ where $A = (-3,1)$ and $B = (2,-2)$.

\require{AMSsymbols} \displaystyle \begin{align} PA &= 2PB \\ PA^2 &= 4 PB^2 \\ (x+3)^2+(y+1)^2 &= 4(x-2)^2+4(y+2)^2 \\ \therefore 3x^2-22x+3y^2+18y+22 &= 0 \end{align}

### Example 9

Find the equation of the locus of a point $P(x,y)$ that moves so that the line $PA$ is perpendicular to line $PB$, where $A = (3,1)$ and $B = (2,5)$.

\require{AMSsymbols} \displaystyle \begin{align} m(PA) \times m(PB) &= -1 \\ \frac{y-1}{x-3} \times \frac{y-5}{x-2} &= -1 \\ (y-1)(y-5) &= -(x-3)(x-2) \\ y^2-6y+5 &= -x^2+5x-6 \\ \therefore x^2+y^2-5x-6y+11 &= 0 \end{align}

## Parabolae as Loci

The equation of the locus of a point $P(x,y)$ that is equidistance from a fixed point, the focus, and fixed line, the directrix.

### Example 10

\require{AMSsymbols} \displaystyle \begin{align} AP &= \sqrt{(x-2)^2+(y+1)^2} \\ BP &= \sqrt{(x-x)^2+(y-3)^2} \\ &= \sqrt{(y-3)^2} \\ &= \left|y-3\right| \\ AP &= BP \\ AP^2 &= BP^2 \\ (x-2)^2+(y+1)^2 &= \left|y-3\right|^2 \\ x^2-4x+4+y^2+2y+1 &= y^2-6y+9 \\ \therefore x^2-4x+8y-4 &= 0 \end{align}

### Example 11

\require{AMSsymbols} \displaystyle \begin{align} AP &= BP \\ AP^2 &= BP^2 \\ AP^2 &= (x-0)^2+(y-1)^2 \\ &=x^2+y^2-2y+1 \\ BP^2 &= (x+2)^2+(y-y)^2 \\ &= x^2+4x+4 \\ x^2+y^2-2y+1 &= x^2+4x+4 &\color{green}{AP^2=BP^2} \\ \therefore y^2-4x-2y-3 &= 0 \end{align} ## Mastering Integration by Parts: The Ultimate Guide

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