સમદ્રીબાજુ ત્રિકોણ ABC (AC = BC) ન | vedclass

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Question

Let ${C}(\alpha, \beta)$

$	herefore(\alpha+2)^{2}+(\beta-3)^{2}=(\alpha-2)^{2}+\beta^{2}$

$\Rightarrow 8 \alpha-6 \beta+9=0$    &nbsp

Vedclass · Accepted Answer

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

Vedclass · Answer

Let ${C}(\alpha, \beta)$

$	herefore(\alpha+2)^{2}+(\beta-3)^{2}=(\alpha-2)^{2}+\beta^{2}$

$\Rightarrow 8 \alpha-6 \beta+9=0$       ......$(1)$

Slope of $\mathrm{AB}=-\frac{3}{4}$

$	herefore$ 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 $\mathrm{C}.$

$	herefore 9 \alpha+12 \beta=43$           .......$(2)$

Solving $( 1)$ and $( 2)$ we have $\alpha=1, \beta=\frac{17}{6}$

$	herefore \quad \mathrm{C} \equiv\left(1, \frac{17}{6}ight)$

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