In a triangle ABC,suppose y=x is the equation | vedclass

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(A) $1$. The angle bisector of $\angle B$ is $y=x$. Since $B(\alpha, \beta)$ lies on this line,we have $\alpha=\beta$. Thus,$B$ is $(\alpha, \alpha)$.

Vedclass · Accepted Answer

$42$

Vedclass · Answer

(A) $1$. The angle bisector of $\angle B$ is $y=x$. Since $B(\alpha, \beta)$ lies on this line,we have $\alpha=\beta$. Thus,$B$ is $(\alpha, \alpha)$.$2$. The side $AC$ has the equation $2x-y=2$. The point $D$ is the intersection of the angle bisector $y=x$ and $AC$ $(2x-y=2)$. Substituting $y=x$ into $2x-y=2$,we get $2x-x=2$,so $x=2$. Thus,$D=(2,2)$.$3$. By the Angle Bisector Theorem in $	riangle ABC$,the bisector of $\angle B$ divides the opposite side $AC$ in the ratio of the adjacent sides: $\frac{AD}{DC} = \frac{AB}{BC}$.$4$. Given $2AB=BC$,we have $\frac{AB}{BC} = \frac{1}{2}$. Therefore,$\frac{AD}{DC} = \frac{1}{2}$.$5$. Using the section formula for point $D(2,2)$ dividing $AC$ in ratio $1:2$,where $A=(4,6)$ and $C=(x_c, y_c)$,we have $2 = \frac{1 \cdot x_c + 2 \cdot 4}{1+2} \implies 6 = x_c + 8 \implies x_c = -2$,and $2 = \frac{1 \cdot y_c + 2 \cdot 6}{1+2} \implies 6 = y_c + 12 \implies y_c = -6$. So $C=(-2,-6)$.$6$. The reflection of $A(4,6)$ about the angle bisector $y=x$ lies on the line $BC$. The reflection $A'$ of $(4,6)$ about $y=x$ is $(6,4)$.$7$. The line $BC$ passes through $B(\alpha, \alpha)$ and $A'(6,4)$. The slope is $m = \frac{\alpha-4}{\alpha-6}$. The equation is $y-\alpha = \frac{\alpha-4}{\alpha-6}(x-\alpha)$.$8$. Since $C(-2,-6)$ lies on $BC$,$-6-\alpha = \frac{\alpha-4}{\alpha-6}(-2-\alpha)$.$9$. Solving this: $(-6-\alpha)(\alpha-6) = (\alpha-4)(-2-\alpha) \implies -(\alpha+6)(\alpha-6) = -(\alpha-4)(\alpha+2) \implies \alpha^2-36 = \alpha^2-2\alpha-8 \implies 2\alpha = 28 \implies \alpha=14$.$10$. Since $\alpha=\beta$,$B=(14,14)$. Then $\alpha+2\beta = 14+2(14) = 14+28 = 42$.

In a $\triangle ABC$,suppose $y=x$ is the equation of the angle bisector of $\angle B$ and the equation of the side $AC$ is $2x-y=2$. If $2AB=BC$ and the points $A$ and $B$ are $(4,6)$ and $(\alpha, \beta)$ respectively,then $\alpha+2\beta$ is equal to

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