# The equation of the three sides of a triangle are $x = 2,y + 1 = 0$ and $x + 2y = 4$. The coordinates of the circumcentre of the triangle are

A. $\left( {4,0} \right)$

B. $\left( {2, - 1} \right)$

C. $\left( {0,4} \right)$

D. $\left( { - 1,2} \right)$

Answer

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**Hint:**Circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle. Here the equations of the three sides of a triangle are given. So, find the vertices of the triangle solving the equations and hence find the equations of any two perpendicular bisectors of two sides of the triangle. Solve these two equations to get the point of intersection of the two bisectors.

**Formula Used**:

The coordinate of the midpoint of the line segment joining the points $A\left( {{x_1},{y_1}} \right)$ and $B\left( {{x_2},{y_2}} \right)$ is given by $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$.

Slope of the line passes through the points $A\left( {{x_1},{y_1}} \right)$ and $B\left( {{x_2},{y_2}} \right)$ is $\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

Equation of a line passes through the points $A\left( {{x_1},{y_1}} \right)$ and having slope $m$ is given by$\dfrac{{y - {y_1}}}{{x - {x_1}}} = m$

If two lines having slopes ${m_1}$ and ${m_2}$ are perpendicular, then product of the slopes is ${m_1}{m_2} = - 1$

**Complete step by step solution:**

First of all find the vertices of the triangle solving the equations of the sides.

The given equations of the three sides of a triangle are $x = 2.....\left( i \right),y + 1 = 0.....\left( {ii} \right)$ and $x + 2y = 4.....\left( {iii} \right)$

Let the triangle be $ABC$ in which the equation of the side $AB$ is $x = 2$, the equation of the side $BC$ is $y + 1 = 0$ and the equation of the side $CA$ is $x + 2y = 4$

Solve the equations of the sides $AB$ and $CA$ to get the coordinates of the point $A$.

Equation of the side $AB$ is given in equation $\left( i \right)$ and equation of the side $CA$ is given in equation $\left( {iii} \right)$.

Putting $x = 2$ in equation $\left( {iii} \right)$, we get

$\begin{array}{l}2 + 2y = 4\\ \Rightarrow 2y = 4 - 2\\ \Rightarrow 2y = 2\\ \Rightarrow y = \dfrac{2}{2} = 1\end{array}$

So, the coordinates of the point $A$ are $\left( {2,1} \right)$

Now, solve the equations of the sides $AB$ and $BC$ to get the coordinates of the point $B$.

Equation of the side $AB$ is given in equation $\left( i \right)$ and equation of the side $BC$ is given in equation $\left( {ii} \right)$.

From equation $\left( {ii} \right)$, we get $y = - 1$

So, the coordinates of the point $B$ are $\left( {2, - 1} \right)$

Now, solve the equations of the sides $BC$ and $CA$ to get the coordinates of the point $C$.

Equation of the side $BC$ is given in equation $\left( {ii} \right)$ and equation of the side $CA$ is given in equation $\left( {iii} \right)$.

From equation $\left( {ii} \right)$, we get $y = - 1$

Putting $y = - 1$ in equation $\left( {iii} \right)$, we get

$\begin{array}{l}x + 2\left( { - 1} \right) = 4\\ \Rightarrow x - 2 = 4\\ \Rightarrow x = 4 + 2\\ \Rightarrow x = 6\end{array}$

So, the coordinates of the point $C$ are $\left( {6, - 1} \right)$

Finally, we get the vertices $A = \left( {2,1} \right),B = \left( {2, - 1} \right),C = \left( {6, - 1} \right)$

Now, find the equations of any two perpendicular bisectors of two sides of the triangle using point-slope form.

Let us find the equation of the perpendicular bisector of the side $BC$.

Slope of the side $BC$ is ${m_1} = \dfrac{{\left( { - 1} \right) - \left( { - 1} \right)}}{{\left( 6 \right) - \left( 2 \right)}} = \dfrac{{ - 1 + 1}}{{6 - 2}} = \dfrac{0}{4} = 0$

If two lines having slopes ${m_1}$ and ${m_2}$ are perpendicular, then product of the slopes is ${m_1}{m_2} = - 1$

So, the slope of the perpendicular bisector of the side $BC$ is ${m_2} = - \dfrac{1}{0}$

Coordinate of the midpoint of the side $BC$ is $\left( {\dfrac{{\left( 2 \right) + \left( 6 \right)}}{2},\dfrac{{\left( { - 1} \right) + \left( { - 1} \right)}}{2}} \right) = \left( {\dfrac{8}{2},\dfrac{{ - 2}}{2}} \right) = \left( {4, - 1} \right)$

We want to find the equation of the perpendicular bisector of the side $BC$, slope of which is ${m_2} = - \dfrac{1}{0}$.

This line passes through the point $\left( {4, - 1} \right)$

Equation of a line passes through the points $A\left( {{x_1},{y_1}} \right)$ and having slope $m$ is given by$\dfrac{{y - {y_1}}}{{x - {x_1}}} = m$

So, equation of the line is $\dfrac{{y - \left( { - 1} \right)}}{{x - \left( 4 \right)}} = - \dfrac{1}{0}$

Simplify the equation of the line.

$\begin{array}{l} \Rightarrow \dfrac{{y + 1}}{{x - 4}} = - \dfrac{1}{0}\\ \Rightarrow x - 4 = 0\\ \Rightarrow x = 4\end{array}$

So, the equation of the perpendicular bisector of the side $BC$ is $x = 4.....\left( {iv} \right)$

Let us find the equation of the perpendicular bisector of the side $CA$.

The slope of the side $CA$ is ${m_3} = \dfrac{{\left( { - 1} \right) - \left( 1 \right)}}{{\left( 6 \right) - \left( 2 \right)}} = \dfrac{{ - 1 - 1}}{{6 - 2}} = \dfrac{{ - 2}}{4} = - \dfrac{1}{2}$

If two lines having slopes ${m_3}$ and ${m_4}$ are perpendicular, then product of the slopes is ${m_3}{m_4} = - 1$

So, the slope of the perpendicular bisector of the side $CA$ is ${m_4} = 2$

Coordinate of the midpoint of the side $CA$ is $\left( {\dfrac{{\left( 2 \right) + \left( 6 \right)}}{2},\dfrac{{\left( 1 \right) + \left( { - 1} \right)}}{2}} \right) = \left( {\dfrac{8}{2},\dfrac{0}{2}} \right) = \left( {4,0} \right)$

We want to find the equation of the perpendicular bisector of the side $CA$, slope of which is ${m_4} = 2$.

This line passes through the point $\left( {4,0} \right)$

Equation of a line passes through the points $C\left( {{x_3},{y_3}} \right)$ and having slope $m$ is given by $\dfrac{{y - {y_3}}}{{x - {x_3}}} = m$

Hence the equation of the perpendicular bisector of the side $CA$ is $\dfrac{{y - 0}}{{x - 4}} = 2$

Arrange this equation as the general form of an equation of a line.

$ \Rightarrow y = 2x - 8.....\left( v \right)$

This is the equation of the perpendicular bisector of the side $CA$.

Now, solve equations $\left( {iv} \right)$ and $\left( v \right)$ to get the point of intersections of these two perpendicular bisectors.

From equations $\left( {iv} \right)$ and $\left( v \right)$, we get

$x = 4$ and $y = 2x - 8$

Put the value of $x$ in equation $\left( v \right)$ to find the value of $y$.

$y = 2 \times 4 - 8 = 8 - 8 = 0$

Finally, we get $x = 4$ and $y = 0$

So, the point of intersection of the two perpendicular bisectors of the side $BC$ and $CA$ is $\left( {4,0} \right)$.

$\therefore $ The center of the circumcircle of $ABC$ is $\left( {4,0} \right)$

**Option ‘A’ is correct**

**Note:**There is another shortcut method to find the circumcentre of a right-angled triangle. In a right-angled triangle, the circumcentre lies at the midpoint of the hypotenuse.

First find the vertices $A = \left( {2,1} \right),B = \left( {2, - 1} \right),C = \left( {6, - 1} \right)$ as done in this solution.

Then find the length of the sides.

Length of side $AB$ is $\left( 1 \right) - \left( { - 1} \right) = 2$ units.

Length of side $BC$ is $6 - 2 = 4$ units.

Length of side $CA$ is $\sqrt {{{\left( {6 - 2} \right)}^2} + {{\left( { - 1 - 1} \right)}^2}} = \sqrt {16 + 4} = \sqrt {20} = 2\sqrt 5 $ units.

Now, $A{B^2} + B{C^2} = 4 + 16 = 20 = C{A^2}$

$\therefore $ The triangle $ABC$ is right angled at $B$.

Since, the circumcentre lies at the midpoint of the hypotenuse in a right-angled triangle.

So, let's find the midpoint of the hypotenuse.

Its hypotenuse is $AC$

Midpoint of $AC$ is $\left( {\dfrac{{\left( 2 \right) + \left( 6 \right)}}{2},\dfrac{{\left( 1 \right) + \left( { - 1} \right)}}{2}} \right) = \left( {\dfrac{8}{2},\dfrac{0}{2}} \right) = \left( {4,0} \right)$

Hence the circle of the circumcircle of the triangle $ABC$ is $\left( {4,0} \right)$.

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