The coordinates of the vertices A and B of an | vedclass

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(B) Let $C$ be $(\alpha, \beta)$.Since $AC = BC$,we have $AC^2 = BC^2$.$(\alpha + 2)^2 + (\beta - 3)^2 = (\alpha - 2)^2 + (\beta - 0)^2$$\alpha^2 + 4\

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$\left(1, \frac{17}{6}\right)$

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(B) Let $C$ be $(\alpha, \beta)$.Since $AC = BC$,we have $AC^2 = BC^2$.$(\alpha + 2)^2 + (\beta - 3)^2 = (\alpha - 2)^2 + (\beta - 0)^2$$\alpha^2 + 4\alpha + 4 + \beta^2 - 6\beta + 9 = \alpha^2 - 4\alpha + 4 + \beta^2$$8\alpha - 6\beta + 9 = 0$  ......$(1)$Slope of $AB = \frac{0 - 3}{2 - (-2)} = -\frac{3}{4}$.The equation of a line parallel to $AB$ is $y = -\frac{3}{4}x + c$.Given $y$-intercept $c = \frac{43}{12}$,the equation is $y = -\frac{3}{4}x + \frac{43}{12}$,which simplifies to $9x + 12y = 43$.Since this line passes through $C(\alpha, \beta)$,we have $9\alpha + 12\beta = 43$ ......$(2)$.Solving equations $(1)$ and $(2)$:From $(1)$,$6\beta = 8\alpha + 9 \Rightarrow 12\beta = 16\alpha + 18$.Substituting into $(2)$: $9\alpha + 16\alpha + 18 = 43$ $\Rightarrow 25\alpha = 25$ $\Rightarrow \alpha = 1$.Then $12\beta = 43 - 9(1) = 34 \Rightarrow \beta = \frac{34}{12} = \frac{17}{6}$.Thus,$C = \left(1, \frac{17}{6}\right)$.

The coordinates of the vertices $A$ and $B$ of an isosceles triangle $ABC$ $(AC = BC)$ are $(-2, 3)$ and $(2, 0)$ respectively. $A$ line parallel to $AB$ and having a $y$-intercept equal to $\frac{43}{12}$ passes through $C$. Then the coordinates of $C$ are:

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