JEE MAIN - Mathematics (2022 - 27th June Morning Shift - No. 10)
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 $$\Delta$$ABC, then 15($$\alpha$$ + $$\beta$$) is equal to :
39
41
51
63
Explanation
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$$\left. \matrix{ 2x + y = 4 \hfill \cr 2x + 6y = 14 \hfill \cr} \right\}y = 2,\,x = 3$$
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 $$15(\alpha + \beta )$$
$$15\left( {{{17} \over 5}} \right) = 51$$
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