Question

The vertices of a triangle are $\left( {a{t_1}{t_2},a\left( {{t_1} + {t_2}} \right)} \right),\left( {a{t_2}{t_3},a\left( {{t_2} + {t_3}} \right)} \right),\left( {a{t_3}{t_1},a\left( {{t_3} + {t_1}} \right)} \right)$. Find the coordinates of its orthocentre.

Answer 1

Hint: The point of intersection of the altitudes of a triangle is known as the orthocentre of the triangle. Here the coordinates of three vertices of a triangle are given. 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:
Let us find the equations of any two altitudes of the triangle using point-slope form.
Let the triangle be $ABC$ in which $A = \left( {a{t_1}{t_2},a\left( {{t_1} + {t_2}} \right)} \right),B = \left( {a{t_2}{t_3},a\left( {{t_2} + {t_3}} \right)} \right),C = \left( {a{t_3}{t_1},a\left( {{t_3} + {t_1}} \right)} \right)$
The values of ${t_1},{t_2},{t_3}$ must not be equal because in that case the points become collinear and hence no triangles will be formed.
Let us find the equation of the altitude through the vertex $A$.
Slope of the side $BC$ is ${m_1} = \dfrac{{\left\{ {a\left( {{t_3} + {t_1}} \right)} \right\} - \left\{ {a\left( {{t_2} + {t_3}} \right)} \right\}}}{{\left( {a{t_3}{t_1}} \right) - \left( {a{t_2}{t_3}} \right)}} = \dfrac{{a{t_3} + a{t_1} - a{t_2} - a{t_3}}}{{a{t_3}{t_1} - a{t_2}{t_3}}} = \dfrac{{a\left( {{t_1} - {t_2}} \right)}}{{a{t_3}\left( {{t_1} - {t_2}} \right)}} = \dfrac{1}{{{t_3}}}$
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 ${m_2} = - {t_3}$
Since, the altitude passes through the vertex $A = \left( {a{t_1}{t_2},a\left( {{t_1} + {t_2}} \right)} \right)$
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 $\dfrac{{y - a\left( {{t_1} + {t_2}} \right)}}{{x - a{t_1}{t_2}}} = - {t_3}$
Simplify the equation of the line.
$ \Rightarrow y - a\left( {{t_1} + {t_2}} \right) = - {t_3}\left( {x - a{t_1}{t_2}} \right)\\ \Rightarrow y - a\left( {{t_1} + {t_2}} \right) = - {t_3}x + a{t_1}{t_2}{t_3}\\ \Rightarrow y = - {t_3}x + a{t_1}{t_2}{t_3} + a\left( {{t_1} + {t_2}} \right)$
So, the equation of the altitude through the point $A$ is $y = - {t_3}x + a{t_1}{t_2}{t_3} + a\left( {{t_1} + {t_2}} \right).....\left( i \right)$
Let us find the equation of the altitude through the vertex $C$.
Slope of the side $AB$ is ${m_3} = \dfrac{{\left\{ {a\left( {{t_2} + {t_3}} \right)} \right\} - \left\{ {a\left( {{t_1} + {t_2}} \right)} \right\}}}{{\left( {a{t_2}{t_3}} \right) - \left( {a{t_1}{t_2}} \right)}} = \dfrac{{a{t_2} + a{t_3} - a{t_1} - a{t_2}}}{{a{t_2}{t_3} - a{t_1}{t_2}}} = \dfrac{{a\left( {{t_3} - {t_1}} \right)}}{{a{t_2}\left( {{t_3} - {t_1}} \right)}} = \dfrac{1}{{{t_2}}}$
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 altitude through the vertex $C$ is ${m_4} = - {t_2}$
Since, the altitude passes through the vertex $C = \left( {a{t_3}{t_1},a\left( {{t_3} + {t_1}} \right)} \right)$
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 $\dfrac{{y - a\left( {{t_3} + {t_1}} \right)}}{{x - a{t_3}{t_1}}} = - {t_2}$
Simplify the equation of the line.
$ \Rightarrow y - a\left( {{t_3} + {t_1}} \right) = - {t_2}\left( {x - a{t_3}{t_1}} \right)\\ \Rightarrow y - a\left( {{t_3} + {t_1}} \right) = - {t_2}x + a{t_1}{t_2}{t_3}\\ \Rightarrow y = - {t_2}x + a{t_1}{t_2}{t_3} + a\left( {{t_3} + {t_1}} \right)$
So, the equation of the altitude through the vertex $C$ is $y = - {t_2}x + a{t_1}{t_2}{t_3} + a\left( {{t_3} + {t_1}} \right).....\left( {ii} \right)$
Now, solve equations $\left( i \right)$ and $\left( {ii} \right)$ to get the point of intersections of these two altitudes.
From equations $\left( i \right)$ and $\left( {ii} \right)$, we get
$ - {t_3}x + a{t_1}{t_2}{t_3} + a\left( {{t_1} + {t_2}} \right) = - {t_2}x + a{t_1}{t_2}{t_3} + a\left( {{t_3} + {t_1}} \right)$
Cancel the term $a{t_1}{t_2}{t_3}$ from both sides.
$ \Rightarrow - {t_3}x + a\left( {{t_1} + {t_2}} \right) = - {t_2}x + a\left( {{t_3} + {t_1}} \right)$
Solve this equation to get the value of $x$
$ \Rightarrow {t_2}x - {t_3}x = a\left( {{t_3} + {t_1}} \right) - a\left( {{t_1} + {t_2}} \right)\\ \Rightarrow \left( {{t_2} - {t_3}} \right)x = a\left( {{t_3} + {t_1} - {t_1} - {t_2}} \right)\\ \Rightarrow \left( {{t_2} - {t_3}} \right)x = a\left( {{t_3} - {t_2}} \right)$
Take out $\left( { - 1} \right)$ as common from the expression of the right hand side.
$ \Rightarrow \left( {{t_2} - {t_3}} \right)x = - a\left( {{t_2} - {t_3}} \right)\\ \Rightarrow x = - \dfrac{{a\left( {{t_2} - {t_3}} \right)}}{{\left( {{t_2} - {t_3}} \right)}}$
, so the term $\left( {{t_2} - {t_3}} \right)$ will be canceled out.
$ \Rightarrow x = - a$
Put the value of $x$ in equation $\left( i \right)$ to find the value of $y$.
$y = - {t_3}\left( { - a} \right) + a{t_1}{t_2}{t_3} + a\left( {{t_1} + {t_2}} \right) = a{t_3} + a{t_1}{t_2}{t_3} + a{t_1} + a{t_2} = a\left( {{t_1} + {t_2} + {t_3} + {t_1}{t_2}{t_3}} \right)$
Finally, we get $x = - a$ and $y = a\left( {{t_1} + {t_2} + {t_3} + {t_1}{t_2}{t_3}} \right)$
So, the point of intersection of the two altitudes is $\left( { - a,a\left( {{t_1} + {t_2} + {t_3} + {t_1}{t_2}{t_3}} \right)} \right)$.

$\therefore $The orthocentre of the triangle $ABC$ is $\left( { - a,a\left( {{t_1} + {t_2} + {t_3} + {t_1}{t_2}{t_3}} \right)} \right)$

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.