Question

the vertices of a triangle are ( 1,2),(h,-3)and(-4,k).if the centroid of the triangle is atbthe point(5,-1) then find the value of root of(h+k)^2+(h+3k)^2.( the answer is 20)

Answer 1

this is your answer
i have used here centriod of triangle formula

Answer 2

The value of $\sqrt{(h+k)^{2}+(h+3k)^{2}}$ is 20.

Step-by-step explanation:

1. We know that centroid of triangle is

  $\bar{X}=\frac{x_{1}+x_{2}+x_{3}}{3}$     ...1)

   $\bar{Y}=\frac{y_{1}+y_{2}+y_{3}}{3}$    ...2)

2. Where

  $(\bar{X},\bar{Y})=(5,-1)$

   $(x_{1},y_{1})=(1,2)$

   $(x_{2},y_{2})=(h,-3)$

   $(x_{3},y_{3})=(-4,k)$

3. Now from equation 1)

  $5=\frac{1+h-4}{3}$

   On solving above equation, we get

   h=18

4. Now from equation 2)

    $-1=\frac{2-3+k}{3}$

    On solving above equation, we get

    k =-2

5. In equation $\sqrt{(h+k)^{2}+(h+3k)^{2}}$ putting the value of h and k

   $\sqrt{(h+k)^{2}+(h+3k)^{2}}$

   $\sqrt{(18-2)^{2}+(18-3×2)^{2}}$  

   $\sqrt{(16)^{2}+(12)^{2}}$  

   = 20