Question

- In a triangle PQR, the co-ordinates of the points \[P\] and \[Q\] are \[\left( { - 2,4} \right)\] and \[\left( {4, - 2} \right)\] respectively. If the equation of the perpendicular bisector of \[PR\] is \[2x - y + 2 = 0\], then the centre of the circumcircle of triangle PQR is
A. \[\left( { - 2, - 2} \right)\]
B. \[\left( {0, - 2} \right)\]
C. \[\left( { - 1,0} \right)\]
D. \[\left( {1,4} \right)\]

Answer 1

-Hint: Circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle. Find the midpoint of the side \[PQ\] and find the slope of the side \[PQ\]. Using these find the equation of the perpendicular bisector of the side \[PQ\] and the equation of the perpendicular bisector of the side \[PR\] is given. Solve these two equations to get the point of intersection of the two bisectors.

Formula used:
The co-ordinate 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:
Find the midpoint of the side \[PQ\], where co-ordinates of the points \[P\] and \[Q\] are \[\left( { - 2,4} \right)\] and \[\left( {4, - 2} \right)\] respectively
Let the midpoint be \[S\]
Then co-ordinate of the midpoint \[S\] is \[\left( {\dfrac{{\left( { - 2} \right) + \left( 4 \right)}}{2},\dfrac{{\left( 4 \right) + \left( { - 2} \right)}}{2}} \right) = \left( {\dfrac{2}{2},\dfrac{2}{2}} \right) = \left( {1,1} \right)\]
Find the equation of the perpendicular bisector of the side \[PQ\].
Slope of the line \[PQ\] is \[\dfrac{{\left( { - 2} \right) - \left( 4 \right)}}{{\left( 4 \right) - \left( { - 2} \right)}} = \dfrac{{ - 2 - 4}}{{4 + 2}} = \dfrac{{ - 6}}{6} = - 1\]
So, the slope of any perpendicular line to the line \[PQ\] is \[\left( { - \dfrac{1}{{ - 1}}} \right)\] i.e. \[1\].
We want to find the equation of the perpendicular bisector of the side \[PQ\], slope of which is \[1\].
This line passes through the point \[S\left( {1,1} \right)\]
So, equation of the line is \[\dfrac{{y - 1}}{{x - 1}} = 1\]
Simplify the equation of the line.
\[\begin{array}{l} \Rightarrow y - 1 = x - 1\\ \Rightarrow y - 1 - x + 1 = 0\\ \Rightarrow y - x = 0\\ \Rightarrow y = x\end{array}\]
So, the equation of the perpendicular bisector of the side \[PQ\] is \[y = x......\left( i \right)\]
and the equation of the perpendicular bisector of the side \[PR\] is \[2x - y + 2 = 0......\left( {ii} \right)\]
Now, solve the equations \[\left( i \right)\] and \[\left( {ii} \right)\].
Substitute \[y = x\] from equation \[\left( i \right)\] in equation \[\left( {ii} \right)\]
\[ \Rightarrow 2x - x + 2 = 0\]
Solve this equation.
\[\begin{array}{l} \Rightarrow x + 2 = 0\\ \Rightarrow x = - 2\end{array}\]
Put the value of \[x\] in equation \[\left( i \right)\]
\[ \Rightarrow y = - 2\]
Finally, we get \[x = - 2\] and \[y = - 2\]
So, the point of intersection of the two perpendicular bisectors of the side \[PQ\] and \[PR\] is \[\left( { - 2, - 2} \right)\].
\[\therefore \]The centre of the circumcircle of triangle is \[\left( { - 2, - 2} \right)\]
Hence option A is correct.

Note: The point of intersection of the perpendicular bisectors of the sides of a triangle is known as the circumcentre of the triangle and the point of intersection of the medians of a triangle is known as the centroid of the triangle. Solving two equations of two straight lines, we obtain the point of intersection of the two straight lines. If there is no solution, then it means that the two lines do not intersect.