Question

Orthocenter of the triangle whose vertices are $\left( {0,0} \right)$, $\left( {2, - 1} \right)$ and $\left( {1,3} \right)$ is:
A) $\left( { - 4, - 1} \right)$
B) $\left( { - \dfrac{4}{7}, - \dfrac{1}{7}} \right)$
C) $\left( {\dfrac{4}{7},\dfrac{1}{7}} \right)$
D) $\left( {4,1} \right)$

Answer 1

Hint:
Here, we are required to find the Orthocenter of a triangle whose vertices are given. We will draw the altitudes of the triangle and find the equation of any two altitudes and eliminating those equations will give us the points of their intersection which is the required points of the Orthocenter of the triangle.

Formulas Used:
Slope of a line, $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
Equation of a line using point-slope form, $y - {y_1} = m\left( {x - {x_1}} \right)$

Complete step by step solution:
Orthocenter is a point where all the three altitudes of a triangle intersect each other.
Now, the slope of a line, $m$ shows the steepness of a line and the distance in which the line is going.
When two lines are perpendicular to each other, i.e. making a right angle, then,
Then the product of their slopes is equal to $ - 1$.
Now, according to the question, make a $\vartriangle ABC$ such that, its vertices are $\left( {0,0} \right)$, $\left( {2, - 1} \right)$ and $\left( {1,3} \right)$respectively.

Now, draw the altitudes of the sides of the $\vartriangle ABC$ as shown in the figure.
As we can see, $AD \bot BC$
Hence,
The product of their slopes will be equal to $ - 1$.
$ \Rightarrow {m_{AD}} \times {m_{BC}} = - 1$
Now, the formula of slope of a line is:
$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$……………………………(1)
Where, $m$is the slope, and $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ are two points on the line.
Hence, slope of the line $BC$, whose two points are $\left( {2, - 1} \right)$ and $\left( {1,3} \right)$.
Hence, substituting $\left( {{x_1},{y_1}} \right) = \left( {2, - 1} \right)$ and $\left( {{x_2},{y_2}} \right) = \left( {1,3} \right)$ in (1), we get,
Slope of $BC$, ${m_{BC}} = \dfrac{{3 - \left( { - 1} \right)}}{{1 - 2}} = \dfrac{{3 + 1}}{{ - 1}} = - 4$
Since, ${m_{AD}} \times {m_{BC}} = - 1$
$ \Rightarrow {m_{AD}} \times \left( { - 4} \right) = - 1$
Dividing both sides by $ - 4$,
$ \Rightarrow {m_{AD}} = \dfrac{1}{4}$
Now, equation of a line using a slope is: $y = mx + c$
Since, slope of the altitude, ${m_{AD}} = \dfrac{1}{4}$
Hence, equation of $AD$ will be: $y = \dfrac{1}{4}x$
Similarly,
$BE \bot AC$
Hence,
The product of their slopes will also be equal to $ - 1$.
$ \Rightarrow {m_{BE}} \times {m_{AC}} = - 1$
Now, slope of the line $AC$, whose two points are $\left( {0,0} \right)$ and $\left( {1,3} \right)$.
Hence, substituting $\left( {{x_1},{y_1}} \right) = \left( {0,0} \right)$ and $\left( {{x_2},{y_2}} \right) = \left( {1,3} \right)$ in (1), we get,
Slope of $AC$, ${m_{AC}} = \dfrac{{3 - 0}}{{1 - 0}} = 3$
Since, ${m_{BE}} \times {m_{AC}} = - 1$
$ \Rightarrow {m_{BE}} \times 3 = - 1$
Dividing both sides by 3,
$ \Rightarrow {m_{BE}} = \dfrac{{ - 1}}{3}$
Now, equation of a line using point-slope is: $y - {y_1} = m\left( {x - {x_1}} \right)$………………..(2)
Since, slope of the altitude, ${m_{BE}} = \dfrac{{ - 1}}{3}$
And the point $B = \left( {2, - 1} \right) = \left( {{x_1},{y_1}} \right)$
Hence, substituting these values in equation (2), we get,
$y - \left( { - 1} \right) = \dfrac{{ - 1}}{3}\left( {x - 2} \right)$
$ \Rightarrow y + 1 = \dfrac{{ - 1}}{3}\left( {x - 2} \right)$
Multiplying both sides by 3,
\[ \Rightarrow 3y + 3 = 2 - x\]
Rewriting this equation,
\[ \Rightarrow x + 3y + 1 = 0\]……………………………………(3)
Hence, this is the equation of the altitude $BE$.
Now, Equation of altitude $AD$ is: $y = \dfrac{1}{4}x$
Or we can write this as: $x = 4y$……………………………………….(4)
Since, the altitudes $AD$ and $BE$ intersect each other at $H$, which is the orthocentre.
Hence, by eliminating the equations of both the altitudes, we will find the required points of their intersection, i.e. point $H$.
Hence, substituting the value of $x$ from equation (4) in (3), we get,
\[ \Rightarrow 4y + 3y + 1 = 0\]
\[ \Rightarrow 7y = - 1\]
Dividing both sides by 7,
\[ \Rightarrow y = - \dfrac{1}{7}\]
Hence, from (4),
$x = 4y$
$ \Rightarrow x = 4\left( {\dfrac{{ - 1}}{7}} \right)$
$ \Rightarrow x = - \dfrac{4}{7}$
Therefore, the Orthocenter of the triangle whose vertices are $\left( {0,0} \right)$, $\left( {2, - 1} \right)$ and $\left( {1,3} \right)$ is:
$\left( {\dfrac{{ - 4}}{7},\dfrac{{ - 1}}{7}} \right)$

Hence, option B is the correct answer.

Note:
An altitude of a triangle is a perpendicular drawn from the vertex of a triangle towards the opposite base. There are three vertices in a triangle; hence, three altitudes can be drawn. The point where all the three altitudes intersect each other is called an Orthocentre. In this question, we should take care while using the formulas as we can use them incorrectly such as: Equation of a line using point-slope form is: $y - {y_1} = m\left( {x - {x_1}} \right)$, here, we can interchange the coordinates and hence, making the equation completely wrong. Hence, such small mistakes should be taken care of.