The centroid G of the triangle with vertices A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) is

G [ (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3 ]

**Question 1 :**

Th e vertices of a triangle are (1, 2), (h, −3) and (−4, k). If the centroid of the triangle is at the point (5, −1) then find the value of √(h + k)^{2} + (h + 3k)^{2}.

**Solution :**

Let A (1, 2) B (h, −3) and C (−4, k) be the vertices of the triangle.

Centroid of the triangle = (5, -1)

G [ (x_{1} + x_{2} + x_{3})/3, (y_{1} + y_{2} + y_{3})/3 ] = (5, -1)

(1 + h - 4)/3, (2 - 3 + k)/3 = (5, -1)

(h - 3)/3, (-1 + k)/3 = (5, -1)

(h - 3)/3 = 5 h - 3 = 15 h = 15 + 3 h = 18 |
(-1 + k)/3 = -1 -1 + k = -3 k = -3 + 1 k = -2 |

√(h + k)^{2} + (h + 3k)^{2 }= √(18 - 2)^{2} + (18 - 6)^{2}

^{ }= √16^{2} + 12^{2}

^{ }= √256 + 144

= √400

= 20

**Question 2 :**

Orthocentre and centroid of a triangle are A(−3, 5) and B(3, 3) respectively. If C is the circumcentre and AC is the diameter of this circle, then find the radius of the circle.

**Solution :**

The distance between the centroid and the orthocenter is twice the distance between the centroid and he circumcenter. The centroid is dividing the othocenter and circumcenter in the ratio 2 : 1.

= (mx_{2} + nx_{1})/(m + n), (my_{2} + ny_{1})/(m + n)

Let C be (a, b)

(2a+1(-3))/(2+1), (2b+1(5))/(2+1) = (3, 3)

(2a-3)/3, (2b+5)/3 = (3, 3)

Equating the x and y coordinates, we get

(2a - 3)/3 = 3 2a - 3 = 9 2a = 9 + 3 2a = 12 a = 6 |
(2b+5)/3 = 3 2b + 5 = 9 2b = 9 - 5 2b = 4 b = 4/2 = 2 |

Hence the circumcenter is (6, 2)

AC = diameter of the circle

A(-3, 5) C(6, 2)

= √(6+3)^{2} + (2-5)^{2}

= √9^{2} + (-3)^{2}

= √(81+9)

= √90

= 3√10

Radius = 3√10/2 = 3√(5/2)

**Question 3 :**

ABC is a triangle whose vertices are A(3, 4), B(−2, −1) and C(5, 3) . If G is the centroid and BDCG is a parallelogram then find the coordinates of the vertex D.

**Solution :**

So, midpoint of diagonal BD = midpoint of diagonal CG

let D = (a, b)

midpoint of BD = [(-2+a)/2, (-1+b)/2 ]

midpoint of CG = [(5+2)/2,(3 + 2)/2 ] = (7/2, 5/2)

hence, [(-2+x)/2, (-1+y)/2] = (7/2, 5/2)

(-2+a)/2 = 7/2 => a = 9

(-1+b)/2 = 5/2 => b = 6

So the required vertex is D (9,6)

**Question 4 :**

If (3/2, 5) (7, -9/2) and (13/2, -13/2) are mid-points of the sides of a triangle, then find the centroid of the triangle.

**Solution :**

Let the given midpoints be (x_{1}, y_{1}) (x_{2}, y_{2}) and (x_{3}, y_{3})

A (x_{1 }+ x_{3 }- x_{2, }y_{1 }+ y_{3 }- y_{2})

B (x_{1 }+ x_{2 }- x_{3, }y_{1 }+ y_{2}_{ }- y_{3})

C (x_{2 }+ x_{3 }- x_{1, }y_{2 }+ y_{3}_{ }- y_{1})

Centroid = (1 + 2 + 12)/3 , (3 + 7 - 16)/3

= (15/3), (-6/3)

= (5, -2)

Hence the required centroid is (5, -2).

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