In an isosceles triangle ABC, the vertex A is (6,1) and the equation of the base BC is 2x+…

Question

In an isosceles triangle $ABC$, the vertex $A$ is $(6,1)$ and the equation of the base $BC$ is $2x+y=4$. Let the point $B$ lie on the line $x+3y=7$. If $(\alpha,\beta)$ is the centroid of $	riangle ABC$, then $15(\alpha+\beta)$ is equal to:

Accepted Answer

$$\left. \begin{pmatrix} 2x + y = 4 \hfill \
  2x + 6y = 14 \hfill \ \end{pmatrix}  \right\}y = 2,\,x = 3$$

B(1, 2)

Let C(k, 4 - 2k)

Now $A{B^2} = A{C^2}$

$${5^2} + {( - 1)^2} = {(6 - k)^2} + {( - 3 + 2k)^2}$$

$$ \Rightarrow 5{k^2} - 24k + 19 = 0$$

$$(5k - 19)(k - 1) = 0 \Rightarrow k = {{19} \over 5}$$

$$C\left( {{{19} \over 5}, - {{18} \over 5}} \right)$$

Centroid ($$\alpha$$, $$\beta$$)

$$\alpha  = {{6 + 1 + {{19} \over 5}} \over 3} = {{18} \over 5}$$

$$\beta  = {{1 + 2 - {{18} \over 5}} \over 3} =  - {1 \over 5}$$

Now $$15(\alpha  + \beta )$$

$$15\left( {{{17} \over 5}} \right) = 51$$

Solutions · 1