Question

Find the orthocentre of the triangle with vertices $\left( {2,\dfrac{{\sqrt 3 - 1}}{2}} \right), \left( {\dfrac{1}{2},\dfrac{{ - 1}}{2}} \right)$ and $\left( {2,\dfrac{{ - 1}}{2}} \right)$.
A. $\left( {\dfrac{3}{2},\dfrac{{\sqrt 3 - 3}}{6}} \right)$
B. $\left( {2,\dfrac{{ - 1}}{2}} \right)$
C. $\left( {\dfrac{5}{4},\dfrac{{\sqrt 3 - 2}}{4}} \right)$
D. $\left( {\dfrac{1}{2},\dfrac{{ - 1}}{2}} \right)$

Answer 1

Hint: First, find the slope of all sides of the given triangle, by using the slope formula. Then by using the relation between the slopes find the type of the triangle. In the end, use the type of triangle to reach the required answer.

Formula Used:
Slope Formula: The slope of the lines passing through the points $\left( {{x_1},{y_1}} \right)$, and $\left( {{x_2},{y_2}} \right)$ is: $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

Complete step by step solution:
The given vertices of the triangle are $\left( {2,\dfrac{{\sqrt 3 - 1}}{2}} \right), \left( {\dfrac{1}{2},\dfrac{{ - 1}}{2}} \right)$ and $\left( {2,\dfrac{{ - 1}}{2}} \right)$.
Let’s calculate the slope of the lines joining the vertices of the triangle.
Let $ABC$ be the triangle with the vertices $A\left( {2,\dfrac{{\sqrt 3 - 1}}{2}} \right), B\left( {\dfrac{1}{2},\dfrac{{ - 1}}{2}} \right)$, and $C\left( {2,\dfrac{{ - 1}}{2}} \right)$.


Image: Triangle $ABC$
The image of a triangle $ABC$ with the vertices $A\left( {2,\dfrac{{\sqrt 3 - 1}}{2}} \right), B\left( {\dfrac{1}{2},\dfrac{{ - 1}}{2}} \right)$, and $C\left( {2,\dfrac{{ - 1}}{2}} \right)$.

Let’s calculate the slopes of the lines joining the vertices of the triangle $ABC$.
For a line $BC$:
Let ${m_1}$ be the slope of the line $BC$.
Apply the slope formula.
${m_1} = \dfrac{{\dfrac{{ - 1}}{2} - \left( {\dfrac{{ - 1}}{2}} \right)}}{{2 - \dfrac{1}{2}}}$
$ \Rightarrow {m_1} = \dfrac{0}{{\left( {\dfrac{3}{2}} \right)}}$
$ \Rightarrow {m_1} = 0$ $.....\left( 1 \right)$

For a line $AB$:
Let ${m_2}$ be the slope of the line $AB$.
Apply the slope formula.
${m_2} = \dfrac{{\dfrac{{\sqrt 3 - 1}}{2} - \left( {\dfrac{{ - 1}}{2}} \right)}}{{2 - \dfrac{1}{2}}}$
$ \Rightarrow {m_2} = \dfrac{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}}{{\left( {\dfrac{3}{2}} \right)}}$
$ \Rightarrow {m_2} = \dfrac{1}{{\sqrt 3 }}$ $.....\left( 2 \right)$

For a line $AC$:
Let ${m_3}$ be the slope of the line $AC$.
Apply the slope formula.
${m_3} = \dfrac{{\dfrac{{\sqrt 3 - 1}}{2} - \left( {\dfrac{{ - 1}}{2}} \right)}}{{2 - 2}}$
$ \Rightarrow {m_3} = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{0}$
$ \Rightarrow {m_3} = \infty $ $.....\left( 3 \right)$

From the equations $\left( 1 \right)$, and $\left( 3 \right)$ we get
The line $BC$ is parallel to the x-axis and the line $AC$ is perpendicular to the x-axis.
So, the lines form a right angle at the point $C$.
Therefore, triangle $ABC$ is a right-angled triangle at the vertex $C\left( {2,\dfrac{{ - 1}}{2}} \right)$.

We know that for a right-angled triangle, the orthocentre is the point where the right angle is formed.
Thus, the coordinates of the orthocentre of the triangle $ABC$ is: $C\left( {2,\dfrac{{ - 1}}{2}} \right)$

Option ‘B’ is correct

Note: To solve these types of questions, an adequate knowledge about the equations of line and slope is required. Using which, the required solution can be obtained.Students should remember the important formulas of finding the equation of line if two points are given and conditions of product of slopes if two lines are perpendicular. Also students often get confused about the slope formula $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$.