CUET UG 2025 — General Test Geometry

The centroid of a triangle with vertices $(x_1, y_1)$(x1​,y1​), $(x_2, y_2)$(x2​,y2​), and $(x_3, y_3)$(x3​,y3​) is:

Centroid $= \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)$=(3x1​+x2​+x3​​,3y1​+y2​+y3​​)

Given:

• First vertex: $(5, 4)$(5,4)
• Second vertex: $(-2, 4)$(−2,4)
• Centroid: $(5, 6)$(5,6)
• Third vertex: $(x_3, y_3)$(x3​,y3​) to be found

---

Using the centroid formula for the $x$x-coordinate:

$$\dfrac{x_1 + x_2 + x_3}{3} = 5$$3x1​+x2​+x3​​=5

$$\dfrac{5 + (-2) + x_3}{3} = 5$$35+(−2)+x3​​=5

$$\dfrac{3 + x_3}{3} = 5$$33+x3​​=5

$$3 + x_3 = 15$$3+x3​=15

$$x_3 = 12$$x3​=12

---

Using the centroid formula for the $y$y-coordinate:

$$\dfrac{y_1 + y_2 + y_3}{3} = 6$$3y1​+y2​+y3​​=6

$$\dfrac{4 + 4 + y_3}{3} = 6$$34+4+y3​​=6

$$\dfrac{8 + y_3}{3} = 6$$38+y3​​=6

$$8 + y_3 = 18$$8+y3​=18

$$y_3 = 10$$y3​=10

---

Therefore, the third vertex is $(12, 10)$(12,10).