Question

Three vertices of a triangle are (h,5),(-4,k),(8,9).

If co-ordinates of the centroid is (-2,6) then find the value of h and k.​

Answer 1

Solution :-

Here, Three vertices of a triangle are

( h, 5 ) , ( -4 , k) and ( 8 , 9 )

Compare these three vertices of a triangle with

(x1 , y1) , ( x2 , y2 ) and ( x3, y3 )

Here, The coordinates of the centroid is

( -2,6 )

Compare these with ( x, y )

By using the centroid ,

x = x1 + x2 + x3 / 3

Put the required values,

-2 = h + ( - 4) + 8 / 3

-2 = h - 4 + 8 / 3

-2 * 3 = h - 4 + 8

-6 = h + 4

h = -6 - 4

h = -10

Now,

y = y1 + y2 + y3 / 3

Put the required values,

6 = 5 + k + 9 / 3

6 * 3 = 5 + k + 9

18 = 14 + k

k = 18 - 14

k = 4

Hence, The value of h and k is

(-10 ) and 4 $.$

Answer 2

$\Large{\underbrace{\underline{\sf{Understanding\; the\; Question}}}}$

Here this is a question from coordinate geometry where we are given coordinates of a triangle and it's centroid. But the major problem is that the coordinates of triangle also have terms h and k, we have to find value of these variables. We can use formula for coordinates of centroid of triangle to find h and k.

So let's start!!

Given coordinates of centroid of  ∆:

• (-2,6)

We have given vertices of ∆:

• A(h,5)

• B(-4,k)

• C(8,9)

Here value of :-

• X1=h

• X2=-4

• X3=8

• Y1=5

• Y2=k

• Y3=9

[You can take these values in any order]

We have formula for centroid of ∆:

$\sf Coordinates \;of \;centroid\:of\:\triangle=\bf\Bigg[\dfrac{X_1+X_2+X_3}{3},\dfrac{Y_1+Y_2+Y_3}{3}\Bigg]$

Now put given values!

$\hookrightarrow \bf (-2,6) =\Bigg[\dfrac{(h)+(-4)+(8)}{3},\dfrac{(5)+(k)+(9)}{3}\Bigg]$

$\hookrightarrow\bf (-2,6) =\Bigg[\dfrac{h-4+8}{3},\dfrac{5+k+9}{3}\Bigg]$

$\hookrightarrow\bf (-2,6) =\Bigg[\dfrac{h+4}{3},\dfrac{14+k}{3}\Bigg]$

_____________________________

Now compare these coordinates!

Firstly we are comparing absicca (value of x axis).

$\hookrightarrow\bf -2 =\dfrac{h+4}{3}$

Now cross multiply!

$\hookrightarrow\bf -2(3) =h+4$

Now solve it by using transportation method for LHS & RHS.

$\hookrightarrow\bf -6 =h+4$

$\hookrightarrow\bf -6-4 =h$

$\underline{\Large{\boxed{\bf -10=h}}}\bigstar$

Now compare ordinate (value of y axis).

$\hookrightarrow\bf 6=\dfrac{14+k}{3}$

Now cross multiply!

$\hookrightarrow\bf 6(3)=14+k$

$\hookrightarrow\bf 18=14+k$

Now transport 14 from RHS to LHS!

$\hookrightarrow\bf 18-14=k$

$\Large{\underline{\boxed{\bf 4=k}}\bigstar$

So the required values of k and h are 4 and -10 respectively.

Additional Information:

$\bigstar\large\boxed{\sf Distance\; formula=\sqrt{(Y_2-Y_1)^2+(X_2-X_1)^2}}$

$\bigstar{\large{\underline{\boxed{\sf Section\: formula=\dfrac{m_1x_2+m_2x_1}{m_1+m_2},\dfrac{m_1y_2+m_2y_1}{m_1+m_2}}}}}$

$\bigstar{\large{\underline{\boxed{\sf Area \:of\:\triangle=\dfrac{1}{2}\Big[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_3+y_1)\Big]}}}}$

$\bigstar{\large{\underline{\boxed{\sf{Mid-point\: formula=\bigg[\dfrac{X_1+X_2}{2},\dfrac{Y_1+Y_2}{2}\bigg]}}}}}$

$\rule{280}{2}$

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