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The co-ordinates 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 co-ordinates of $C$ are :-
$\left( { - \frac{3}{4},1} \right)$
$\left( {1,\frac{{17}}{6}} \right)$
$\left( {\frac{2}{3},\frac{4}{5}} \right)$
$(1, 0)$
Solution
Let ${C}(\alpha, \beta)$
$\therefore(\alpha+2)^{2}+(\beta-3)^{2}=(\alpha-2)^{2}+\beta^{2}$
$\Rightarrow 8 \alpha-6 \beta+9=0$ ……$(1)$
Slope of $\mathrm{AB}=-\frac{3}{4}$
$\therefore$ Equation of line parallel to $AB$ and having
$y$ -intercept $=\frac{43}{12}$ is
$y=-\frac{3}{4} x+\frac{43}{12} \Rightarrow 9 x+12 y=43$
$\because$ It passes through $\mathrm{C}.$
$\therefore 9 \alpha+12 \beta=43$ …….$(2)$
Solving $( 1)$ and $( 2)$ we have $\alpha=1, \beta=\frac{17}{6}$
$\therefore \quad \mathrm{C} \equiv\left(1, \frac{17}{6}\right)$
