#### Example – 01:

- A(3, 1) and B(2, -5) are the vertices of a triangle. Find the equation of locus of the third vertex C, if the centroid of triangle lies on the locus y = 3 + 2x
^{2}. **Solution:**

let G(h, k) be the centroid. It lies on the locus y = 3 + 2x^{2}.

Hence it should satisfy the equation of the locus.

k = 3 + 2h^{2} ……….. (1)

Let C(x’, y’) be the third vertex of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

(-4 + y’)/3 = 3 + 2((5+x’)/3)^{2}

(-4 + y’)/3 = 3 + 2(25+ 10x’ +x’^{2})/9

Multiplying both sides by 9

3(-4 + y’) = 27 + 2(25+ 10x’ +2x’^{2})

-12 + 3y’ = 27 + 50+ 20x’ + 2x’^{2}

27 + 50+ 20x’ + 2x’^{2 }+ 12 – 3y’ = 0

2x’^{2} + 20x’ – 3y’ + 89 = 0

Replacing x’ with x and y’ with y we get

2x^{2} + 20x – 3y + 89 = 0

**Ans:** The equation of the required locus is 2x^{2} + 20x – 3y + 89 = 0

#### Example – 02:

- A(3, -2) and B(-2, 4) are the vertices of a triangle. Find the equation of locus of the third vertex C, if the centroid of triangle lies on the locus 3x
^{2}– 5y^{2}= 15. **Solution:**

let G(h, k) be the centroid. It lies on the locus 3x^{2} – 5y^{2} = 15

Hence it should satisfy the equation of the locus.

3h^{2} – 5k^{2} = 15 ……….. (1)

Let C(x’, y’) be the third vertex of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

3((1+ x’)/3)^{2} – 5((2 + y’)/3)^{2} = 15

3((1+ 2x’ + x’^{2})/9) – 5((4 + 4y’ + y’^{2})/9) = 15

Multiplying both sides by 9

3(1+ 2x’ + x’^{2}) – 5(4+ 4y’ + y’^{2}) = 135

3+ 6x’ + 3x’^{2} – 20 – 20y’ – 5y’^{2 }– 135 = 0

3x’^{2} – 5y’^{2 }+ 6x’ – 20y’ – 152 = 0

Replacing x’ with x and y’ with y we get

3x^{2} – 5y^{2 }+ 6x – 20y – 152 = 0

**Ans:** The equation of the required locus is 3x^{2} – 5y^{2 }+ 6x – 20y – 152 = 0

#### Example – 03:

- A(2, -5) and B(-7, 6) are the vertices of a triangle. Find the equation of locus of the third vertex C, if the centroid of triangle lies on the locus 3x – 4y = 11.
**Solution:**

let G(h, k) be the centroid. It lies on the locus 3x – 4y + 11 = 0

Hence it should satisfy the equation of the locus.

3h – 4k + 11 = 0 ……….. (1)

Let C(x’, y’) be the third vertex of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

3((-5+ x’)/3) – 4((1 + y’)/3) + 11 = 0

Multiplying both sides by 3

3(-5+ x’) – 4(1 + y’) + 33 = 0

– 15+ 3x’ – 4 – 4y’ + 33 = 0

3x’ – 4y’ – 19 + 33 = 0

3x’ – 4y’ + 14 = 0

Replacing x’ with x and y’ with y we get

3x – 4y + 14 = 0

**Ans:** The equation of the required locus is 3x – 4y + 14 = 0

#### Example – 03:

- A(5, -2) and B(2, 4) are the vertices of a triangle. The third vertex C of triangle lies on the locus y = 1 + x + x
^{2}. Find the equation of locus of the centroid of the triangle. **Solution:**

let C(h, k) be the centroid. It lies on the locus y = 1 + x + x^{2}

Hence it should satisfy the equation of the locus.

k = 1 + h + h^{2} ……….. (1)

Let G(x’, y’) be the centroid of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

(3y’ – 2) = 1 + (3x’ – 7) + (3x’ – 7)^{2}

3y’ – 2 = 1 + 3x’ – 7 + 9x’^{2 }– 42x’ + 49

1 + 3x’ – 7 + 9x’^{2 }– 42x’ + 49 – 3y’ + 2 = 0

9x’^{2 }– 39x’ – 3y’ + 45 = 0

3x’^{2 }– 13x’ – y’ + 15 = 0

Replacing x’ with x and y’ with y we get

3x^{2 }– 13x – y + 15 = 0

**Ans:** The equation of the required locus is 3x^{2 }– 13x – y + 15 = 0

#### Example – 04:

- A(3, -2) and B(-2, 4) are the vertices of a triangle. The third vertex C of triangle lies on the locus y = 2 + 3x + x
^{2}. Find the equation of locus of the centroid of the triangle. **Solution:**

let C(h, k) be the centroid. It lies on the locus y = 2 + 3x + x^{2}

Hence it should satisfy the equation of the locus.

k = 2 + 3h + h^{2} ……….. (1)

Let G(x’, y’) be the centroid of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

3y’ – 7 = 2 + 3(3x’ – 2) + (3x’ – 2)^{2}

3y’ – 7 = 2 + 9x’ – 6 + (9x’^{2} – 12x’ + 4)

2 + 9x’ – 6 + 9x’^{2} – 12x’ + 4 – 3y’ + 7 = 0

9x’^{2 }– 3x’ – 3y’ + 7 = 0

Replacing x’ with x and y’ with y we get

9x^{2 }– 3x – 3y + 7 = 0

**Ans:** The equation of the required locus is 9x^{2 }– 3x – 3y + 7 = 0

#### Example – 05:

- A(1, 4) and B(6, -3) are the vertices of a triangle. The third vertex C of triangle lies on the locus 5x + 4y = 9. Find the equation of locus of the centroid of the triangle.
**Solution:**

let C(h, k) be the centroid. It lies on the locus 5x + 4y = 9

Hence it should satisfy the equation of the locus.

5h + 4k = 9 ……….. (1)

Let G(x’, y’) be the centroid of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

5(3x’ – 7) + 4(3y’ – 1) = 9

15x’ – 35 + 12y’ – 4 = 9

15x’ + 12y’ -39 – 9 = 0

15x’ + 12y’ – 48 = 0

5x’ + 4y’ – 16 = 0

Replacing x’ with x and y’ with y we get

5x + 4y – 16 = 0

**Ans:** The equation of the required locus is 5x + 4y – 16 = 0

#### Example – 06:

- A(4, -3) and C(0, 2) are the vertices of a triangle. The third vertex B of triangle lies on the locus y = 1+ x
^{2}. Find the equation of locus of the centroid of the triangle. **Solution:**

let B(h, k) be the centroid. It lies on the locus y = 1+ x^{2}

Hence it should satisfy the equation of the locus.

k = 1+ h^{2} ……….. (1)

Let G(x’, y’) be the centroid of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

(3y’ + 1) = 1 +(3x’ – 4)^{2}

3y’ + 1 = 1 +9x’^{2} – 24x’ + 16

1 +9x’^{2} – 24x’ + 16 – 3y’ – 1 = 0

9x’^{2} – 24x’ – 3y’ + 16 = 0

Replacing x’ with x and y’ with y we get

9x^{2} – 24x – 3y + 16 = 0

**Ans:** The equation of the required locus is 9x^{2} – 24x – 3y + 16 = 0