Question

Find the sum of coordinates of the orthocentre of the triangle whose sides are x = 3, y = 4, 3x + 4y = 6.
(a) (0, 0)
(b) (3, 0)
(c) (0, 4)
(d) (3, 4)

Answer 1

Hint: We will take the help of the formula $y-{{y}_{1}}=n\left( x-{{x}_{1}} \right)$ to find the equation of the lines where n is called the slope. Also, we will use the concept of the slopes that if m is a slope of one line and n is the slope of another line then their product is going to be equal to – 1. Thus we get $\left( n \right)\left( m \right)=-1$. These formulas will help us to solve the question and lead to the right answer.

Complete step-by-step answer:
The diagram for the question is shown below.

The points D, G and H are found out by solving the given equations. For example, by solving x = 3 and 3x + 4y = 6 we will have 3(3) + 4y = 6. This further results into, 4y = 6 – 9. Therefore the value of y is $\dfrac{-3}{4}$. So, the point G is $\left( 3,\dfrac{-3}{4} \right)$. We can also write it in decimals as G(3, - 0.75). Similarly we will get D(3, 4) and H(-3.3, 4).
First we will understand the concept of orthocentre. The orthocentre is the point where all altitudes of the triangle meet. Here the orthocentre is represented by the point J. Let us suppose that the point J is denoted by the point (x,y). To find the orthocentre of the triangle we will find the slopes of the line DE and the line HG. Now we will find the slope of the line 3x + 4y = 6 which is basically, HG. As we know that we can also write the equation 3x + 4y = 6 as y = - 0.75x + 1.5. By comparing it to the equation of the line which is given by y = mx + c, we will get that m = - 0.75. So, the slope of the line HG is - 0.75.
As we can clearly see that the line DE is perpendicular to HG. If we consider the slope of DE to be n and the slope of HG is m therefore, the product of the two slopes m and n are going to be equal to – 1. Thus we get
$\begin{align}
  & \left( n \right)\left( m \right)=-1 \\
 & \Rightarrow \left( n \right)=\dfrac{-1}{m} \\
 & \Rightarrow \left( n \right)=\dfrac{-1}{-0.75} \\
 & \Rightarrow \left( n \right)=1.33 \\
\end{align}$
So, the slope of DE is 1.33. Now we will use the formula $y-{{y}_{1}}=n\left( x-{{x}_{1}} \right)$ to find the equation of DE. Thus, by substituting points of D(3,4) as $\left( {{x}_{1}},{{y}_{1}} \right)$ we will have
$\begin{align}
  & y-4=1.33\left( x-3 \right) \\
 & \Rightarrow y-4=\left( x\left( 1.33 \right)-3\left( 1.33 \right) \right) \\
 & \Rightarrow y-4=1.33x-3.99 \\
 & \Rightarrow 1.33x-y-3.99+4=0 \\
 & \Rightarrow 1.33x-y+0.01=0...........(i) \\
\end{align}$
The slope of DG can be carried out by the formula $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ where $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 3,4 \right),\left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,-0.75 \right)$. Thus we get
$\begin{align}
  & m=\dfrac{-0.75-4}{3-3} \\
 & \Rightarrow m=\infty \\
\end{align}$
Now as the line HI is perpendicular to DG therefore we will use the product formula of slopes. If n be the slope of HI and m is the slope of DG so, we will have
$\begin{align}
  & \left( n \right)\left( m \right)=-1 \\
 & \Rightarrow \left( n \right)=\dfrac{-1}{\infty } \\
 & \Rightarrow \left( n \right)=0 \\
\end{align}$
So, the slope of HI is 0. Now we will use the formula $y-{{y}_{1}}=n\left( x-{{x}_{1}} \right)$ to find the equation of DE. Thus, by substituting points of H(-3.3, 4) as $\left( {{x}_{1}},{{y}_{1}} \right)$ we will have
$\begin{align}
  & y-4=0\left( x+3.3 \right) \\
 & \Rightarrow y-4=0 \\
 & \Rightarrow y=4 \\
\end{align}$
Now we will substitute the value in equation (i). Therefore, we will get
$\begin{align}
  & 1.33x-4+0.01=0 \\
 & \Rightarrow 1.33x-3.99=0 \\
 & \Rightarrow 1.33x=3.99 \\
 & \Rightarrow x=\dfrac{3.99}{1.33} \\
 & \Rightarrow x=3 \\
\end{align}$
Therefore, the orthocentre of the point J(x, y) is (3,4). Hence the correct option is (d).

Note: Don’t be confused about the orthocentre as the centre of the triangle. Alternatively we can solve it as by finding the slope of DH can be carried out by substituting the points of D(3, 4) and H(-3.3, 4) in the equation $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ we will get
$\begin{align}
  & m=\dfrac{4-4}{-3.3-3} \\
 & \Rightarrow 6.3m=0 \\
 & \Rightarrow m=0 \\
\end{align}$
So, the slope of DH is 0. As the lines DH and GF are perpendicular to each other therefore, the slope n of GF is
$\begin{align}
  & \left( m \right)\left( n \right)=-1 \\
 & \Rightarrow n=\dfrac{-1}{0} \\
 & \Rightarrow n=\infty \\
\end{align}$
Now we will find the equation of the line GF by using the formula $y-{{y}_{1}}=\infty \left( x-{{x}_{1}} \right)$ and in this we will substitute the point G(3, - 0.75) as $\left( {{x}_{1}},{{y}_{1}} \right)$ will result into
$\begin{align}
  & y-\left( -0.75 \right)=\infty \left( x-3 \right) \\
 & \Rightarrow \dfrac{y+0.75}{\infty }=\left( x-3 \right) \\
 & \Rightarrow x-3=0 \\
 & \Rightarrow x=3 \\
\end{align}$
We will substitute the value of y in equation (i). Thus we get
$\begin{align}
  & 3+3y-15=0 \\
 & \Rightarrow 3y-12=0 \\
 & \Rightarrow 3y=12 \\
 & \Rightarrow y=4 \\
\end{align}$
Hence, the point of the orthocentre J is given by (3, 4) and the correct option is (d).
Moreover, there is another method also which is solved below.
We can also solve this question alternative by finding the slope of the line JE. For this we will first find the value of the point E. This can be found out by using the midpoint formula. If we consider the points of H as (a,b) and G as (c,d) then the midpoint E will be
$\begin{align}
  & \left( \dfrac{c+a}{2},\dfrac{b+d}{2} \right)=\left( \dfrac{3+\left( -3.3 \right)}{2},\dfrac{4+\left( -0.75 \right)}{2} \right) \\
 & \Rightarrow \left( \dfrac{c+a}{2},\dfrac{b+d}{2} \right)=\left( \dfrac{-0.3}{2},\dfrac{3.25}{2} \right) \\
 & \Rightarrow \left( \dfrac{c+a}{2},\dfrac{b+d}{2} \right)=\left( -0.15,1.625 \right) \\
\end{align}$
Thus the point E is (- 0.15, 1.625). Now we will find the slope of the JE. This is given by $\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}={{m}_{2}}\left( \dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{2}}} \right)$. By considering the point $\left( {{x}_{1}},{{y}_{1}} \right)=\left( h,k \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( -0.15,1.625 \right)$ we will get
$\begin{align}
  & \dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}={{m}_{2}}\left( \dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right) \\
 & \Rightarrow \dfrac{y-k}{1.625-k}={{m}_{2}}\left( \dfrac{x-h}{-0.15-h} \right) \\
 & \Rightarrow {{m}_{2}}=\dfrac{\dfrac{y-k}{1.625-k}}{\dfrac{x-h}{-0.15-h}} \\
\end{align}$
As we can clearly see that the lines JE and HG are perpendicular to each other. Therefore, the product of the slopes is going to be equal to – 1. Thus we get
$\begin{align}
  & m\times {{m}_{2}}=-1 \\
 & \Rightarrow \left( 3 \right)\left( \dfrac{\dfrac{y-k}{1.625-k}}{\dfrac{x-h}{-0.15-h}} \right)=-1 \\
 & \Rightarrow 3\left( \dfrac{y-k}{1.625-k} \right)=-\left( \dfrac{x-h}{-0.15-h} \right) \\
 & \Rightarrow \left( \dfrac{3y-3k}{1.625-k} \right)=\left( \dfrac{-x+h}{-0.15-h} \right) \\
 & \Rightarrow \left( 3y-3k \right)\left( -0.15-h \right)=\left( -x+h \right)\left( 1.625-k \right) \\
 & \Rightarrow \left( 3y\left( -0.15-h \right)-3k\left( -0.15-h \right) \right)=\left( -x\left( 1.625-k \right)+h\left( 1.625-k \right) \right) \\
 & \Rightarrow -0.45y-3hy+0.45k+3kh=-1.625x+kx+1.625h-hk \\
 & \Rightarrow -0.45y-3hy+0.45k+3kh+hk=-1.625x+kx+1.625h \\
 & \Rightarrow -0.45y-3hy+0.45k+4hk=-1.625x+kx+1.625h.......................(i) \\
\end{align}$
Similarly we will do this with the lines of JI and DG and solve as usual.