Question

What is the circumcentre of the triangle formed by the lines \[y = x,y = 2x\] and \[y = 3x + 4\]?
A. \[\left( {6,8} \right)\]
B. \[\left( {6, - 8} \right)\]
C. \[\left( {3,4} \right)\]
D. \[\left( { - 3, - 4} \right)\]

Answer 1

Hint: Circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle. Find the coordinates of the vertices of the triangle by solving the equations and hence find the coordinates of the midpoints of any two sides of the triangle and find the slope of the two sides. Using these find the equation of the perpendicular bisector of the two sides. Solve these two equations to get the point of intersection of the two perpendicular 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:
Let \[ABC\] be the triangle, equations of the three sides of which are given.
Let equations of the sides \[AB,BC,CA\] of the triangle \[ABC\] be \[y = x - - - - - \left( i \right),y = 2x - - - - - \left( {ii} \right),y = 3x + 4 - - - - - \left( {iii} \right)\] respectively.
Let us solve the equations pairwise to get the coordinates of the vertices \[A,B,C\].
From equations \[\left( i \right)\] and \[\left( {ii} \right)\], we get
\[x = 2x\]
\[ \Rightarrow x - 2x = 0\]
\[ \Rightarrow - x = 0\]
\[ \Rightarrow x = 0\]
Putting \[x = 0\] in equation \[\left( i \right)\], we get \[y = 0\]
So, the coordinates of the vertex \[B\] i.e. the point of intersection of the two lines \[\left( i \right)\] and \[\left( {ii} \right)\] are \[\left( {0,0} \right)\]
From equations \[\left( {ii} \right)\] and \[\left( {iii} \right)\], we get
\[2x = 3x + 4\]
\[ \Rightarrow 2x - 3x = 4\]
\[ \Rightarrow - x = 4\]
\[ \Rightarrow x = - 4\]
Putting \[x = - 4\] in equation \[\left( {ii} \right)\], we get \[y = 2 \times \left( { - 4} \right) = - 8\]
So, the coordinates of the vertex \[C\] i.e. the point of intersection of the two lines \[\left( {ii} \right)\] and \[\left( {iii} \right)\] are \[\left( { - 4, - 8} \right)\]
From equations \[\left( i \right)\] and \[\left( {iii} \right)\], we get
\[x = 3x + 4\]
\[ \Rightarrow x - 3x = 4\]
\[ \Rightarrow - 2x = 4\]
\[ \Rightarrow 2x = - 4\]
\[ \Rightarrow x = - 2\]
Putting \[x = - 2\] in equation \[\left( i \right)\], we get \[y = - 2\]
So, the coordinates of the vertex \[A\] i.e. the point of intersection of the two lines \[\left( i \right)\] and \[\left( {iii} \right)\] are \[\left( { - 2, - 2} \right)\]
Thus, the vertices of the triangle \[ABC\] are \[A\left( { - 2, - 2} \right),B = \left( {0,0} \right),C = \left( { - 4, - 8} \right)\].
Find the midpoints of the sides \[AB\] and \[BC\]
The co-ordinates of the midpoint of the side \[AB\] is \[\left( {\dfrac{{\left( { - 2} \right) + \left( 0 \right)}}{2},\dfrac{{\left( { - 2} \right) + \left( 0 \right)}}{2}} \right) = \left( {\dfrac{{ - 2}}{2},\dfrac{{ - 2}}{2}} \right) = \left( { - 1, - 1} \right)\]
The coordinates of the midpoint of the side \[BC\] is \[\left( {\dfrac{{\left( 0 \right) + \left( { - 4} \right)}}{2},\dfrac{{\left( 0 \right) + \left( { - 8} \right)}}{2}} \right) = \left( {\dfrac{{ - 4}}{2},\dfrac{{ - 8}}{2}} \right) = \left( { - 2, - 4} \right)\]
Find the equations of the perpendicular bisectors of the sides \[AB\] and \[BC\].
Slope of the side \[AB\] is \[\dfrac{{\left( 0 \right) - \left( { - 2} \right)}}{{\left( 0 \right) - \left( { - 2} \right)}} = \dfrac{{0 + 2}}{{0 + 2}} = \dfrac{2}{2} = 1\]
So, the slope of any perpendicular line to the line \[AB\] is \[\left( { - 1} \right)\]
We want to find the equation of the perpendicular bisector of the side \[AB\], slope of which is \[\left( { - 1} \right)\].
This line passes through the point \[\left( { - 1, - 1} \right)\]
So, equation of the line is \[\dfrac{{y - \left( { - 1} \right)}}{{x - \left( { - 1} \right)}} = - 1\]
Simplify the equation of the line.
\[ \Rightarrow \dfrac{{y + 1}}{{x + 1}} = - 1\]
\[ \Rightarrow y + 1 = - x - 1\]
\[ \Rightarrow y = - x - 1 - 1\]
\[ \Rightarrow y = - x - 2\]
So, the equation of the perpendicular bisector of the side \[AB\] is \[y = - x - 2 - - - - - \left( {iv} \right)\]
Slope of the side \[BC\] is \[\dfrac{{\left( { - 8} \right) - \left( 0 \right)}}{{\left( { - 4} \right) - \left( 0 \right)}} = \dfrac{{ - 8 - 0}}{{ - 4 - 0}} = \dfrac{{ - 8}}{{ - 4}} = 2\]
So, the slope of any perpendicular line to the line \[BC\] is \[\left( { - \dfrac{1}{2}} \right)\]
We want to find the equation of the perpendicular bisector of the side \[BC\], slope of which is \[\left( { - \dfrac{1}{2}} \right)\].
This line passes through the point \[\left( { - 2, - 4} \right)\]
So, equation of the line is \[\dfrac{{y - \left( { - 4} \right)}}{{x - \left( { - 2} \right)}} = - \dfrac{1}{2}\]
Simplify the equation of the line.
\[ \Rightarrow \dfrac{{y + 4}}{{x + 2}} = - \dfrac{1}{2}\]
\[ \Rightarrow y + 4 = - \dfrac{1}{2}\left( {x + 2} \right)\]
\[ \Rightarrow y + 4 = - \dfrac{1}{2}x - 1\]
\[ \Rightarrow y = - \dfrac{1}{2}x - 1 - 4\]
\[ \Rightarrow y = - \dfrac{1}{2}x - 5\]
So, the equation of the perpendicular bisector of the side \[BC\] is \[y = - \dfrac{1}{2}x - 5 - - - - - \left( v \right)\]
Now, solve the equations \[\left( {iv} \right)\] and \[\left( v \right)\] to find the coordinates of the point of intersection.
From equations \[\left( {iv} \right)\] and \[\left( v \right)\], we get
\[ - x - 2 = - \dfrac{1}{2}x - 5\]
Solve this equation to find the value of \[x\]
Take out the minus sign from both the sides.
\[ \Rightarrow x + 2 = \dfrac{1}{2}x + 5\]
\[ \Rightarrow x - \dfrac{1}{2}x = 5 - 2\]
\[ \Rightarrow \dfrac{1}{2}x = 3\]
\[ \Rightarrow x = 6\]
Put the value of \[x\] in equation \[\left( {iv} \right)\] to find the value of \[y\].
\[ \Rightarrow y = - 6 - 2 = - 8\]
Finally, we get \[x = 6\] and \[y = - 8\]
So, the point of intersection of the two perpendicular bisectors of the side \[AB\] and \[BC\] is \[\left( {6, - 8} \right)\].
\[\therefore \]The centre of the circumcircle of the triangle \[ABC\] is \[\left( {6, - 8} \right)\]
Hence option B is correct.

Note: To solve this type of problem students should know that 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.