Finding Equation of Locus by Locus Definition

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.

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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

\( \require{AMSsymbols} \displaystyle \begin{align} AP &= BP \\ AP^2 &= BP^2 \\ AP^2 &= (x-0)^2+(y-1)^2 \\ &=x^2+y^2-2y+1 \\ BP^2 &= (x+2)^2+(y-y)^2 \\ &= x^2+4x+4 \\ x^2+y^2-2y+1 &= x^2+4x+4 &\color{green}{AP^2=BP^2} \\ \therefore y^2-4x-2y-3 &= 0 \end{align} \)

 

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