Question

Find it :-
if the coordinates of the circumcentre of the triangle formed by the lines $3x-2y=6,3x+4y+12=0$ and $3x-8y+12=0$ is $(\alpha ,\beta )$ then $3\left(\beta -\alpha \right)$equals

Answer 1

$L_1\equiv 3x-2y-6=0$
$L_2\equiv 3x+4y+12=0$
$L_3\equiv 3x-8y+12=0$
Simultaneously solving the equations $L_1$ and $L_2$ gives one vertex $\left(A\right)$ as $(0,-3)$.
SImilarly, solving the equations $L_2$ and $L_3$ gives second vertex $\left(B\right)$ as $(-4,0)$.
Finally, solving the equations $L_3$ and $L_1$ gives third vertex $\left(C\right)$ as $(4,3)$.
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Circumcentre is point of interesection of perpendicular bisectors of all the three points.
Let us first find the perpendicular bisector of $AB$:
Mid-point of $AB\equiv (\frac{0-4}{2},\frac{-3+0}{2})\equiv (-2,-\frac{3}{2})$
Slope of $AB=\frac{y_2-y_1}{x_2-x_1}=\frac{0-(-3)}{-4-0}=-\frac{3}{4}$
Therefore, slope of line perpenicular to $AB=\frac{4}{3}$ (Since product of slopes of perpendicular lines = -1)
Therefore, equation of bisector:
$(y+\frac{3}{2})=\frac{4}{3}(x+2)⟹8x-6y+7=0$ (1)
(1) is the required perpendicular bisector.
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Now, finding the perpendicular bisector of $BC$
Mid-point of $BC=(0,\frac{3}{2})$
Slope of $BC=\frac{3}{8}$
Therefore, slope of line perpenicular to $AB=-\frac{8}{3}$
Therefore, equation of bisector:
$y-\frac{3}{2}=-\frac{8}{3}(x-0)⟹16x+6y-9=0$ (2)
(2) is the required perpendicular bisector.
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Simultaneously solving (1) and (2) gives:
$x=\frac{1}{12},y=\frac{23}{18}$
$\therefore \beta =\frac{23}{18},\alpha =\frac{1}{12}$
$⟹3(\beta -\alpha )=3(\frac{23}{18}-\frac{1}{12})=3(\frac{46-3}{36})=\frac{43}{12}$