Question

In which quadrant does the orthocentre of the triangle formed by the lines $x + y = 1,2x + 3y = 6$ and $4x - y + 4 = 0$ lie?
A. $1$st
B. $2$nd
C. $3$rd
D. $4$th

Answer 1

Hint: The point of intersection of the altitudes of a triangle is known as the orthocentre of the triangle. Here the equations of the sides of a triangle are given. Solve the equations to get the coordinates of the vertices. Then find the equations of any two altitudes of the triangle using point-slope form and solve it to get the point of intersection.

Formula Used:
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 $x + y = 1.....\left( i \right),2x + 3y = 6.....\left( {ii} \right)$ and $4x - y = - 4.....\left( {iii} \right)$
Let the triangle be $ABC$ in which the equation of the side $AB$ is $x + y = 1$, the equation of the side $BC$ is $2x + 3y = 6$ and the equation of the side $CA$ is $4x - y = - 4$
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 $\left( {iii} \right)$.
Adding the equations $\left( i \right)$ and $\left( {iii} \right)$, we get $5x = - 3 \Rightarrow x = - \dfrac{3}{5}$
Putting $x = - \dfrac{3}{5}$ in equation $\left( i \right)$, we get
$\left( { - \dfrac{3}{5}} \right) + y = 1\\ \Rightarrow y = 1 + \dfrac{3}{5}\\ \Rightarrow y = \dfrac{8}{5}$
So, the coordinates of the point $A$ are $\left( { - \dfrac{3}{5},\dfrac{8}{5}} \right)$
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 $\left( {ii} \right)$.
Multiplying both sides of equation $\left( i \right)$ by $3$, we get $3x + 3y = 3.....\left( {iv} \right)$
Subtracting equation $\left( {ii} \right)$ from equation $\left( {iv} \right)$, we get $x = - 3$
Putting $x = - 3$ in equation $\left( i \right)$, we get
$\left( { - 3} \right) + y = 1\\ \Rightarrow y = 1 + 3\\ \Rightarrow y = 4$
So, the coordinates of the point $B$ are $\left( { - 3,4} \right)$
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 $\left( {iii} \right)$.
Multiplying both sides of equation $\left( {iii} \right)$ by $3$, we get $12x - 3y = - 12.....\left( v \right)$
Adding equations $\left( {ii} \right)$ and equation $\left( v \right)$, we get $14x = - 6 \Rightarrow x = - \dfrac{6}{{14}} = - \dfrac{3}{7}$
Putting $x = - \dfrac{3}{7}$ in equation $\left( {ii} \right)$, we get
$2 \times \left( { - \dfrac{3}{7}} \right) + 3y = 6\\ \Rightarrow - \dfrac{6}{7} + 3y = 6\\ \Rightarrow 3y = 6 + \dfrac{6}{7}\\ \Rightarrow 3y = \dfrac{{48}}{7}\\ \Rightarrow y = \dfrac{{48}}{7} \times \dfrac{1}{3}\\ \Rightarrow y = \dfrac{{16}}{7}$
So, the coordinates of the point $C$ are $\left( { - \dfrac{3}{7},\dfrac{{16}}{7}} \right)$
Finally, we get the vertices $A = \left( { - \dfrac{3}{5},\dfrac{8}{5}} \right),B = \left( { - 3,4} \right),C = \left( { - \dfrac{3}{7},\dfrac{{16}}{7}} \right)$
Now, find the equations of any two altitudes of the triangle using point-slope form.
Let us find the equation of the altitude through the vertex $A$.
Slope of the side $BC$ is ${m_1} = \dfrac{{\left( {\dfrac{{16}}{7}} \right) - \left( 4 \right)}}{{\left( { - \dfrac{3}{7}} \right) - \left( { - 3} \right)}} = \dfrac{{\left( {\dfrac{{16 - 28}}{7}} \right)}}{{\left( {\dfrac{{ - 3 + 21}}{7}} \right)}} = \dfrac{{\left( { - \dfrac{{12}}{7}} \right)}}{{\left( {\dfrac{{18}}{7}} \right)}} = - \dfrac{{12}}{7} \times \dfrac{7}{{18}} = - \dfrac{2}{3}$
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 altitude through the vertex $A$ is ${m_2} = \dfrac{3}{2}$
Since, this line passes through the vertex $A\left( { - \dfrac{3}{5},\dfrac{8}{5}} \right)$
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 altitude is $\dfrac{{y - \left( {\dfrac{8}{5}} \right)}}{{x - \left( { - \dfrac{3}{5}} \right)}} = \dfrac{3}{2}$
Simplify the equation of the line.
$ \Rightarrow \dfrac{{y - \dfrac{8}{5}}}{{x + \dfrac{3}{5}}} = \dfrac{3}{2}\\ \Rightarrow \dfrac{{\left( {\dfrac{{5y - 8}}{5}} \right)}}{{\left( {\dfrac{{5x + 3}}{5}} \right)}} = \dfrac{3}{2}\\ \Rightarrow \dfrac{{5y - 8}}{{5x + 3}} = \dfrac{3}{2}\\ \Rightarrow 2\left( {5y - 8} \right) = 3\left( {5x + 3} \right)\\ \Rightarrow 10y - 16 = 15x + 9\\ \Rightarrow 10y = 15x + 9 + 16\\ \Rightarrow 10y = 15x + 25$
Take out $5$ as common from both sides.
$ \Rightarrow 2y = 3x + 5\\ \Rightarrow y = \dfrac{{3x + 5}}{2}.....\left( {vi} \right)$
So, the equation of the altitude through the point $A$ is $2y = 3x + 5$
Let us find the equation of the altitude through the vertex $C$.
Slope of the side $AB$ is ${m_3} = \dfrac{{\left( {\dfrac{8}{5}} \right) - \left( 4 \right)}}{{\left( { - \dfrac{3}{5}} \right) - \left( { - 3} \right)}} = \dfrac{{\left( {\dfrac{{8 - 20}}{5}} \right)}}{{\left( {\dfrac{{ - 3 + 15}}{5}} \right)}} = \dfrac{{\left( { - \dfrac{{12}}{5}} \right)}}{{\left( {\dfrac{{12}}{5}} \right)}} = - \dfrac{{12}}{5} \times \dfrac{5}{{12}} = - 1$
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 altitude through the vertex $C$ is ${m_4} = 1$
Since, this line passes through the vertex $C\left( { - \dfrac{3}{7},\dfrac{{16}}{7}} \right)$
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 altitude is $\dfrac{{y - \left( {\dfrac{{16}}{7}} \right)}}{{x - \left( { - \dfrac{3}{7}} \right)}} = 1$
Simplify the equation of the line.
$ \Rightarrow \dfrac{{y - \dfrac{{16}}{7}}}{{x + \dfrac{3}{7}}} = 1\\ \Rightarrow \dfrac{{\left( {\dfrac{{7y - 16}}{7}} \right)}}{{\left( {\dfrac{{7x + 3}}{7}} \right)}} = 1\\ \Rightarrow \dfrac{{7y - 16}}{{7x + 3}} = 1\\ \Rightarrow 7y - 16 = 7x + 3\\ \Rightarrow 7y = 7x + 3 + 16\\ \Rightarrow 7y = 7x + 19\\ \Rightarrow y = \dfrac{{7x + 19}}{7}......\left( {vii} \right)$
So, the equation of the altitude through the vertex $C$ is $7y = 7x + 19$
Now, solve equations $\left( {vi} \right)$ and $\left( {vii} \right)$ to get the point of intersections of these two perpendicular bisectors.
From equations $\left( {vi} \right)$ and $\left( {vii} \right)$, we get
$\dfrac{{3x + 5}}{2} = \dfrac{{7x + 19}}{7}$
Solve this equation to get the value of $x$
By cross-multiplication, we get
$ \Rightarrow 7\left( {3x + 5} \right) = 2\left( {7x + 19} \right)\\ \Rightarrow 21x + 35 = 14x + 38\\ \Rightarrow 21x - 14x = 38 - 35\\ \Rightarrow 7x = 3\\ \Rightarrow x = \dfrac{3}{7}$
Put the value of $x$ in equation $\left( {vi} \right)$ to find the value of $y$.
$y = \dfrac{{3 \times \left( {\dfrac{3}{7}} \right) + 5}}{2} = \dfrac{{\left( {\dfrac{9}{7} + 5} \right)}}{2} = \left( {\dfrac{{9 + 35}}{7}} \right) \times \dfrac{1}{2} = \dfrac{{44}}{7} \times \dfrac{1}{2} = \dfrac{{22}}{7}$
Finally, we get $x = \dfrac{3}{7}$ and $y = \dfrac{{22}}{7}$
So, the point of intersection of the two altitudes is $\left( {\dfrac{3}{7},\dfrac{{22}}{7}} \right)$.
$\therefore $The orthocentre of the triangle $ABC$ is $\left( {\dfrac{3}{7},\dfrac{{22}}{7}} \right)$
Clearly, the orthocentre lies in $1$st quadrant, abscissa and ordinate both being positive.

Option ‘A’ is correct

Note: Many students get confused about the orthocentre, circumcentre and the centroid. You should remember that the point of intersection of the altitudes of a triangle is known as orthocentre of the triangle and the point of intersection of the medians of a triangle is known as centroid of the triangle and 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 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.