Question

A(h, -6), B(2, 3) and C(-6, k) are the co-ordinates of vertices of a triangle

whose centroid is G (1, 5). Find h.​

Answer 1

Required Answer:-

What is Centroid? And how to determine the medians when the coordinates of the vertices is given?

The centroid is the intersection of medians in a triangle. In coordinate geometry, we can find the mean of x coordinates and y coordinates of the vertices of a ∆ respectively to get the X and y coordinate of the centroid$.$

Solving the above,

By using the above concept,

(x1 + x2 + x3 / 3, y1 + y2 + y3 / 3) = (x' , y')

Where,

• (x1, y1), (x2,y2) and (x3,y3) are the coordinates of the vertices of the triangle.
• (x' , y') is the coordinate of the centroid.

Then, ATQ:

⇒ (h + 2 - 6/3, -6 + 3 + k/3) = (1,5)

⇒ (h - 4/3, -3 + k/3) = (1,5)

Comparing individually, we just need 'h':

⇒ h - 4/3 = 1

⇒ h - 4 = 3.

⇒ h = 7

$\therefore$ The required value of h is 7 (Ans)

Answer 2

Given :-

Points

• A(h,-6)
• B(2,3)
• C(-6,k)
• G(1,5)

To Find

Value of h

Solution :-

We know that

$\sf \bigg(\dfrac{x1 + x2 + x3} 3\bigg),\bigg( \dfrac{y1 + y2 + y3}{ 3}\bigg) = (x' , y')$

Now

⇒ (h + 2 - 6/3, -6 + 3 + k/3) = (1,5)

⇒(h + -4/3, -6 + 3 k/3) = (1,5)

⇒ (h - 4/3, -3 + k/3) = (1,5)

Now

Lets solve them separately

h - 4/3 = 1                         -3 + k/3 = 5

h - 4 = (3)(1)                       -3 + k = (5)(3)

h - 4 = 3                            -3 + k = 15

h = 4 + 3                            k = -3 + 15

h = 7                                  k = 18