Equations of Loci

7.4 Equations of Loci
 
The image is an educational graphic titled ‘Equations of Loci.’ It includes the logo of ‘Pandai’ at the top. Below the title, there are two sections connected by dashed arrows. The first section, labeled ‘Locus (plural: Loci),’ defines a locus as ‘The set of all points that satisfy a particular condition or a set of conditions.’ The second section, labeled ‘Equation of a Locus,’ defines it as ‘The mathematical representation that describes the position of all such points.’ The overall design uses a combination of blue, white, and red colors.
 
Equation of Locus

Locus of moving point \(P(x,y)\) which distance is constant from a single fixed point \(A(x_1,y_1)\) is:

\((x-x_1)^2+(y-y_1)^2=r^2\)


Locus of moving point \(P(x,y)\) which distance from two fixed points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is always constant in the ratio \(m:n\) is: 

\(\dfrac{(x-x_1)^2+(y-y_1)^2}{(x-x_2)^2+(y-y_2)^2}=\dfrac{m^2}{n^2}\)


Locus of moving point \(P(x,y)\) which distance from two fixed points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is the same is:

\((x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2\)

 
Example \(1\)
Question

Find the equation of the locus of a moving point \(P(x,y)\) so that its distance is always \(5\) units from a fixed point \(Q(2,4)\).

Solution

The formula involved is (a fixed point):

\((x-x_1)^2+(y-y_1)^2=r^2\)


Substitute the value of the coordinate \(Q(2,4)\) and the distance of \(5\) units into the formula:

\((x-2)^2+(y-4)^2=5^2\)


Expand and simplify the resulting equation:

\(x^2-4x+4+y^2-8y+16=25 \\ x^2+y^2-4x-8y-5=0\)


The equation of the locus of a moving point \(P\) is:

\(x^2+y^2-4x-8y-5=0\)

 
Example \(2\)
Question

Find the equation of the locus of the moving point \(P(x,y)\) so that its distance from point \(A(-2,3)\) and point \(B(4,-1)\) is the same.

Solution

The formula involved is (two fixed point):

\((x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2\)


Substitute the value of the coordinates \(A(-2,3)\) and \(B(4,-1)\) into the formula:

\((x-(-2))^2+(y-3)^2=(x-4)^2+(y-(-1))^2\)


Expand and simplify the resulting equation:

\((x+2)^2+(y-3)^2=(x-4)^2+(y+1)^2 \\ x^2+2x+4+y^2-6y+9=x^2-8x+16+y^2+2y+1 \\ 10x-8y-4=0\)


The equation of the locus of a moving point \(P\) is:

\(10x-8y-4=0\)

 
Example \(3\)
Question

\(A(2,0)\) and \(B(0,-2)\) are two fixed points. Point \(P\) moves in a ratio such that \(AP:PB=1:2\). Find the equation of the locus of the moving point \(P\).

Solution

The formula involved is (two fixed point, ratio):

\(\dfrac{(x-x_1)^2+(y-y_1)^2}{(x-x_2)^2+(y-y_2)^2}=\dfrac{m^2}{n^2}\)


Substitute the value of the coordinates \(A(2,0)\) and \(B(4,-1)\) with ratio \(AP:PB=1:2\) into the formula:

\(\dfrac{AP}{PB}=\dfrac{1}{2}\)

\(\dfrac{(x-2)^2+(y-0)^2}{(x-0)^2+(y+2)^2}=\dfrac{1}{2^2}\)


Expand and simplify the resulting equation:

\(4[(x-2)^2+y^2]=x^2+(y+2)^2 \\ 4(x^2-4x+4)+4y^2=x^2+(y^2+4y+4) \\ 4x^2-16x+16+4y^2=x^2+y^2+4y+4 \\ 3x^2+3y^2-16x-4y+12=0\)


The equation of the locus of a moving point \(P\) is:

\(3x^2+3y^2-16x-4y+12=0\)

 

 

Equations of Loci

7.4 Equations of Loci
 
The image is an educational graphic titled ‘Equations of Loci.’ It includes the logo of ‘Pandai’ at the top. Below the title, there are two sections connected by dashed arrows. The first section, labeled ‘Locus (plural: Loci),’ defines a locus as ‘The set of all points that satisfy a particular condition or a set of conditions.’ The second section, labeled ‘Equation of a Locus,’ defines it as ‘The mathematical representation that describes the position of all such points.’ The overall design uses a combination of blue, white, and red colors.
 
Equation of Locus

Locus of moving point \(P(x,y)\) which distance is constant from a single fixed point \(A(x_1,y_1)\) is:

\((x-x_1)^2+(y-y_1)^2=r^2\)


Locus of moving point \(P(x,y)\) which distance from two fixed points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is always constant in the ratio \(m:n\) is: 

\(\dfrac{(x-x_1)^2+(y-y_1)^2}{(x-x_2)^2+(y-y_2)^2}=\dfrac{m^2}{n^2}\)


Locus of moving point \(P(x,y)\) which distance from two fixed points \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is the same is:

\((x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2\)

 
Example \(1\)
Question

Find the equation of the locus of a moving point \(P(x,y)\) so that its distance is always \(5\) units from a fixed point \(Q(2,4)\).

Solution

The formula involved is (a fixed point):

\((x-x_1)^2+(y-y_1)^2=r^2\)


Substitute the value of the coordinate \(Q(2,4)\) and the distance of \(5\) units into the formula:

\((x-2)^2+(y-4)^2=5^2\)


Expand and simplify the resulting equation:

\(x^2-4x+4+y^2-8y+16=25 \\ x^2+y^2-4x-8y-5=0\)


The equation of the locus of a moving point \(P\) is:

\(x^2+y^2-4x-8y-5=0\)

 
Example \(2\)
Question

Find the equation of the locus of the moving point \(P(x,y)\) so that its distance from point \(A(-2,3)\) and point \(B(4,-1)\) is the same.

Solution

The formula involved is (two fixed point):

\((x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2\)


Substitute the value of the coordinates \(A(-2,3)\) and \(B(4,-1)\) into the formula:

\((x-(-2))^2+(y-3)^2=(x-4)^2+(y-(-1))^2\)


Expand and simplify the resulting equation:

\((x+2)^2+(y-3)^2=(x-4)^2+(y+1)^2 \\ x^2+2x+4+y^2-6y+9=x^2-8x+16+y^2+2y+1 \\ 10x-8y-4=0\)


The equation of the locus of a moving point \(P\) is:

\(10x-8y-4=0\)

 
Example \(3\)
Question

\(A(2,0)\) and \(B(0,-2)\) are two fixed points. Point \(P\) moves in a ratio such that \(AP:PB=1:2\). Find the equation of the locus of the moving point \(P\).

Solution

The formula involved is (two fixed point, ratio):

\(\dfrac{(x-x_1)^2+(y-y_1)^2}{(x-x_2)^2+(y-y_2)^2}=\dfrac{m^2}{n^2}\)


Substitute the value of the coordinates \(A(2,0)\) and \(B(4,-1)\) with ratio \(AP:PB=1:2\) into the formula:

\(\dfrac{AP}{PB}=\dfrac{1}{2}\)

\(\dfrac{(x-2)^2+(y-0)^2}{(x-0)^2+(y+2)^2}=\dfrac{1}{2^2}\)


Expand and simplify the resulting equation:

\(4[(x-2)^2+y^2]=x^2+(y+2)^2 \\ 4(x^2-4x+4)+4y^2=x^2+(y^2+4y+4) \\ 4x^2-16x+16+4y^2=x^2+y^2+4y+4 \\ 3x^2+3y^2-16x-4y+12=0\)


The equation of the locus of a moving point \(P\) is:

\(3x^2+3y^2-16x-4y+12=0\)