Question

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

Answer 1

Hint: In this question, we first need to find the vertices of the triangle from the given line equations. Then as we already know that circumcentre is equidistant from all the vertices by using the distance formula we can find the coordinates of the circumcentre.

Complete step-by-step answer:
$\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$

DISTANCE FORMULA: Distance between two points (x1 , y1) and (x2 , y2) , is
$\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
Now, to find the vertices of the triangle we need to solve the given line equations.
Let us consider the line equations
$3x-3y=6$
$3x+4y+12=0$
Now, on subtracting these two equations we get,
$\Rightarrow -3y-4y-12=6$
Let us now rearrange the terms

& \Rightarrow 7y=-18 \\\ & \therefore y=\dfrac{-18}{7} \\\ \end{aligned}$$ Now, by substituting this value of y in any of the two equations we get value of x. $$\begin{aligned} & \Rightarrow 3x-3\left( \dfrac{-18}{7} \right)=6 \\\ & \Rightarrow 3x=6-\dfrac{54}{7} \\\ \end{aligned}$$ $$\Rightarrow 3x=-\dfrac{12}{7}$$ Now, on dividing both sides with 3 we get, $$\therefore x=-\dfrac{4}{7}$$ Now, on considering other two line equations $$\begin{aligned} & 3x+4y+12=0 \\\ & 3x-8y+12=0 \\\ \end{aligned}$$ Let us now subtract these two equations then we get, $$\Rightarrow 4y+8y=0$$ Now, on further simplification we get, $$\therefore y=0$$ Now, by substituting this value of y in any of the two equations we get value of x. $$\begin{aligned} & \Rightarrow 3x+4\left( 0 \right)+12=0 \\\ & \Rightarrow 3x=-12 \\\ \end{aligned}$$ Let us now divide both sides with 3. $$\therefore x=-4$$ Now, on considering other two line equations $$3x-3y=6$$ $$3x-8y+12=0$$ Now, on subtracting these two equations we get, $$\Rightarrow -3y+8y-12=6$$ Let us now rearrange the terms $$\Rightarrow 5y=18$$ Now, on dividing both sides with 5 we get, $$\therefore y=\dfrac{18}{5}$$ Now, by substituting this value of y in any of the two equations we get value of x. $$\begin{aligned} & \Rightarrow 3x-3\left( \dfrac{18}{5} \right)=6 \\\ & \Rightarrow 3x=6+\dfrac{54}{5} \\\ & \Rightarrow 3x=\dfrac{84}{5} \\\ \end{aligned}$$ Now, on dividing both sides with 3 we get, $$\therefore x=\dfrac{28}{5}$$ Now, the coordinates of the given triangle are $$\left( \dfrac{-4}{7},\dfrac{-18}{7} \right),\left( -4,0 \right),\left( \dfrac{28}{5},\frac{18}{5} \right)$$ Given, that the circumcentre of the triangle, denoted by S in the figure below, is: $$\left( \alpha ,\beta \right)$$ As we already know that circumcenter is equidistant from the vertices we get, Distance between circumcentre and the first vertex is: $$\begin{aligned} & \Rightarrow \sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \\\ & \Rightarrow \sqrt{{{\left( \alpha -\left( \dfrac{-4}{7} \right) \right)}^{2}}+{{\left( \beta -\left( \frac{-18}{7} \right) \right)}^{2}}} \\\ & \Rightarrow \sqrt{{{\left( \alpha +\dfrac{4}{7} \right)}^{2}}+{{\left( \beta +\dfrac{18}{7} \right)}^{2}}} \\\ \end{aligned}$$ Now, let us calculate the distance between the circumcentre and the second vertex. $$\begin{aligned} & \Rightarrow \sqrt{{{\left( \alpha -\left( -4 \right) \right)}^{2}}+{{\left( \beta -0 \right)}^{2}}} \\\ & \Rightarrow \sqrt{{{\left( \alpha +4 \right)}^{2}}+{{\left( \beta \right)}^{2}}} \\\ \end{aligned}$$ Let us now calculate the distance between the third vertex and the circumcentre. $$\Rightarrow \sqrt{{{\left( \alpha -\dfrac{28}{5} \right)}^{2}}+{{\left( \beta -\dfrac{18}{5} \right)}^{2}}}$$ By equating these distances we get, $$\Rightarrow \sqrt{{{\left( \alpha +\dfrac{4}{7} \right)}^{2}}+{{\left( \beta +\dfrac{18}{7} \right)}^{2}}}=\sqrt{{{\left( \alpha +4 \right)}^{2}}+{{\left( \beta \right)}^{2}}}=\sqrt{{{\left( \alpha -\dfrac{28}{5} \right)}^{2}}+{{\left( \beta -\dfrac{18}{5} \right)}^{2}}}$$$$$$ Now, on equating the first two terms we get, $$\Rightarrow \sqrt{{{\left( \alpha +\dfrac{4}{7} \right)}^{2}}+{{\left( \beta +\dfrac{18}{7} \right)}^{2}}}=\sqrt{{{\left( \alpha +4 \right)}^{2}}+{{\left( \beta \right)}^{2}}}$$ Now, by squaring on both sides and expanding the terms we get, $$\Rightarrow {{\alpha }^{2}}+{{\left( \dfrac{4}{7} \right)}^{2}}+2\times \alpha \times \dfrac{4}{7}+{{\beta }^{2}}+{{\left( \dfrac{18}{7} \right)}^{2}}+2\times \alpha \times \dfrac{18}{7}={{\alpha }^{2}}+16+8\alpha +{{\beta }^{2}}$$ Now, by cancelling the common terms on both the sides and on further simplification we get, $$\Rightarrow \dfrac{16}{49}+\dfrac{8}{7}\alpha +\dfrac{324}{49}+\dfrac{36}{7}\beta =16+8\alpha $$ Now, on rearranging the terms we get, $$\begin{aligned} & \Rightarrow \dfrac{8}{7}\alpha -8\alpha +\dfrac{36}{7}\beta =16-\dfrac{340}{49} \\\ & \Rightarrow \dfrac{-48}{7}\alpha +\dfrac{36}{7}\beta =\dfrac{444}{49} \\\ \end{aligned}$$ Now, by dividing by 12 and multiplying with 7 on both sides we get, $$\Rightarrow -4\alpha +3\beta =\dfrac{37}{7}...........\left( 1 \right)$$ Similarly, by solving the other two equations we get, $$\Rightarrow \sqrt{{{\left( \alpha +4 \right)}^{2}}+{{\left( \beta \right)}^{2}}}=\sqrt{{{\left( \alpha -\dfrac{28}{5} \right)}^{2}}+{{\left( \beta -\dfrac{18}{5} \right)}^{2}}}$$ Now, by squaring on both sides and expanding the terms we get, $$\Rightarrow {{\alpha }^{2}}+{{\left( \dfrac{28}{5} \right)}^{2}}+2\times \alpha \times \dfrac{28}{5}+{{\beta }^{2}}+{{\left( \dfrac{18}{5} \right)}^{2}}+2\times \beta \times \dfrac{18}{5}={{\alpha }^{2}}+16+8\alpha +{{\beta }^{2}}$$ Now, by cancelling the common terms on both the sides and on further simplification we get, $$\Rightarrow \frac{784}{25}+\dfrac{54}{5}\alpha +\dfrac{324}{25}+\dfrac{36}{5}\beta =16+8\alpha $$ Now, on rearranging the terms we get, $$\begin{aligned} & \Rightarrow \dfrac{54}{5}\alpha -8\alpha +\dfrac{36}{5}\beta =16-\dfrac{1108}{25} \\\ & \Rightarrow \dfrac{14}{5}\alpha +\frac{36}{5}\beta =\dfrac{-708}{25} \\\ \end{aligned}$$ Now, on dividing by 2 and multiplying by 5 on both sides we get, $$\Rightarrow 7\alpha +18\beta =\dfrac{-354}{5}...........\left( 2 \right)$$ Now, on solving the equations (1) and (2) we get, Let us multiply the equation (1) with 6 then we get, $$\begin{aligned} & \Rightarrow -24\alpha +18\beta =\dfrac{222}{7} \\\ & \Rightarrow 7\alpha +18\beta =\dfrac{-354}{5} \\\ \end{aligned}$$ Now, on subtracting these two equations we get, $$\Rightarrow -31\alpha =\dfrac{222}{7}+\dfrac{354}{5}$$ $$\therefore \alpha =\dfrac{-3588}{1085}$$ Now, by substituting this value in any of the above two equations we get, $$\begin{aligned} & \Rightarrow 3\beta =\dfrac{37}{7}+4\left( \dfrac{-3588}{1085} \right) \\\ & \therefore 3\beta =\dfrac{-8617}{1085} \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow 3\left( \beta -\alpha \right)=\dfrac{-8617}{1085}-3\left( \dfrac{-3588}{1085} \right) \\\ & \Rightarrow 3\left( \beta -\alpha \right)=\dfrac{10764-8617}{1085} \\\ & \therefore 3\left( \beta -\alpha \right)=\dfrac{2147}{1085} \\\ \end{aligned}$$ Note:Instead of finding the vertices by finding the point of intersection of the lines and then from that using the distance formula to get the coordinates of the circumcentre. We can find the equation of the circle touching the three straight lines and then find its circumcentre from that.It is important to note that there should be no calculation errors or the missing terms because there are so many equations included there is a chance for missing out the terms which results in the change of the coordinates and then changing the final result.