Question

Select the correct option: -
Orthocenter of a triangle made by the lines. \[x+y+1=0,\,x-y+3=0,\,2x+y=7\,\]is.
(p) (-4,-7)
(q) (-7,11)
(r) (1, -2)
(s) (-1,2)
(t) (4, -7)

Answer 1

Hint: We have given the three lines. First, we find the vertices of the triangle formed by the given three lines. After that find the equation of two altitudes drawn from the two different vertices of the triangle formed by the given three lines. The equation of the altitudes can be written in point slope form by using the vertex as point. And slope can be formed as altitude is perpendicular to the side.

Complete step-by-step solution -
Find the intersection of two altitudes and name that point as orthocentre, since, orthocentre of a triangle is actually the point of intersection of the altitudes of the triangle .
We have given equation of three lines,
\[x+y+1=0,\,\,x-y+3=0,\,\,2x+y-7=0\,\]
 

Now, by simultaneously solving equations:
$x-y+3=0$ and $x+y+1=0$, we will get vertex B.
$\begin{align}
  & \therefore x-y+3=0 \\
 & \Rightarrow x=y-3...........(i) \\
\end{align}$
Putting $x=(y-3)$ in equation $\left( x+y+1=0 \right)$, we will get
$\Rightarrow y-3+y+1=0$
$\begin{align}
  & 2y-2=0 \\
 & \Rightarrow 2y=2 \\
 & \therefore y=1 \\
\end{align}$
Put y=1 in equation (i), we get
$\begin{align}
  & x=1-3=(-2) \\
 & \therefore vertex\,B\,\equiv \left( -2,1 \right) \\
\end{align}$
Now by simultaneously solving equation $x+y+1=0$ and $2x+y-7=0$, we will get vertex c
$\begin{align}
  & \therefore x+y+1=0 \\
 & \Rightarrow y=-1-x......(ii) \\
\end{align}$
Putting $y=-1-x$ in equation $2x+y-7=0$, we will get
$\begin{align}
  & \Rightarrow 2x+\left( -1-x \right)-7=0 \\
 & \Rightarrow 2x-1-x-7=0 \\
 & \Rightarrow x-8=0 \\
 & \Rightarrow x=8 \\
\end{align}$
Put x=8 in equation (ii), we get
$\begin{align}
  & y=-1-8=-9 \\
 & \therefore vertex\,C=\left( 8.-9 \right) \\
\end{align}$
Now by simultaneously solving equations
$x-y+3=0\,and\,2x+y-7=0$, we will get vertex A
$\begin{align}
  & \therefore x-y+3=0 \\
 & \Rightarrow x=y-3........(iii) \\
\end{align}$
Putting $x=\left( y-3 \right)$ in equation $2x+y-7=0$, we will get
$\begin{align}
  & \Rightarrow 2\left( y-3 \right)+y-7=0 \\
 & \Rightarrow 2y-6+y-7=0 \\
 & \Rightarrow 3y-13=0 \\
 & \Rightarrow 3y=13 \\
 & \Rightarrow y=\dfrac{13}{3} \\
\end{align}$
Putting $y=\dfrac{13}{3}$ in equation (iii), we get
$x=\dfrac{13}{3}-3$
$x=\dfrac{13-9}{3}=\dfrac{4}{3}$
$\therefore Vertex\,A=\left( \dfrac{4}{3},\dfrac{13}{3} \right)$
Now, we will find the equation of altitude BE, which is perpendicular to side AC and passes through point B.
We know, slope of side AC can be found as-
$A\left( \dfrac{4}{3},\dfrac{13}{3} \right),\,C\left( 8,-9 \right)$
Slope of AC= $\dfrac{-9-\dfrac{13}{3}}{8-\dfrac{4}{3}}$ = $\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)$
$=\dfrac{-\dfrac{40}{3}}{\dfrac{20}{3}}=-2$
Since, $BE\bot AC$
$\therefore $slope of AC × slope of BE= (-1)
(-2) × Slope of BE= (-1)
$\therefore $Slope of BE= $\dfrac{1}{2}$
Hence, equation of BE can be written in point slope form,
$y-{{y}_{0}}=m\left( x-{{x}_{0}} \right)$
$\begin{align}
  & \Rightarrow y-1=\dfrac{1}{2}\left( x-\left( -2 \right) \right) \\
 & \Rightarrow 2\left( y-1 \right)=x+2 \\
 & \Rightarrow 2y-2=x+2 \\
 & \Rightarrow 2y-x-2-2=0 \\
 & \Rightarrow 2y-x-4=0 \\
 & \Rightarrow x-2y+4=0...........(iv) \\
 & \left[ here,\,\left( {{x}_{0}},{{y}_{0}} \right)\equiv B\equiv \left( -2,1 \right) \right] \\
\end{align}$
Now, we will find the equation of altitude CF, which is perpendicular to the side AB and passes through point C.
Slope of side AB can be found as-
$A\left( \dfrac{4}{3},\dfrac{13}{3} \right);B\left( -2,1 \right)$
Slope of AB =
$\begin{align}
& \left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right) \\
  & \dfrac{1-\dfrac{13}{3}}{-2-\dfrac{4}{3}} \\
 & =\dfrac{-\dfrac{10}{3}}{-\dfrac{10}{3}}=1 \\
 \end{align}$
Since, $CF\bot AB$
$\therefore $ Slope of AB × Slope of CF = (-1)
1 × slope of CF = (-1)
$\therefore $Slope of CF = (-1)
Hence, equation of CF can be written in point slope form, $y-{{y}_{0}}=m\left( x-{{x}_{0}} \right)$
\[\begin{align}
  & \Rightarrow y-\left( -9 \right) = \left( -1 \right)\left( x-8 \right) \\
 & \Rightarrow y+9=-x+8 \\
 & \Rightarrow y+x+9-8=0 \\
 & \Rightarrow x+y+1=0..............(v) \\
\end{align}\]
Now,
We will find the intersection of altitudes CF and BE by solving equation (iv) and (v).
From equation (iv) and (v)-
$\begin{align}
  & x-2y+4=0 \\
 & \Rightarrow x=2y-4.............(vii) \\
\end{align}$
Putting x=2y-4 in equation (v), we get-
$\begin{align}
  & \Rightarrow 2y-4+y+1=0 \\
 & \Rightarrow 3y-3=0 \\
 & \Rightarrow y=1 \\
\end{align}$
Put y=1 in equation (vii), we get
$\begin{align}
  & \therefore x=2(1)-4 \\
 & \therefore x=(-2) \\
 & \therefore Point\,of\,intersection\,=(-2, 1) \\
\end{align}$

Note: Students can easily solve this question in this way: -
Since slope of equation \[ax+by+c=0=\dfrac{-a}{b}\]
We see clearly that slope of AB = $\dfrac{-1}{-1}=1$
AB: $x+y+1=04$
$\therefore slope\,of\,BC=\dfrac{-1}{1}=(-1)$
We notice that,
Slope of AB × Slope of BC = 1×(-1)=-1
This means AB is perpendicular to BC. Therefore, $\vartriangle ABC$ is a right-angled triangle, with right angle at B.

Now, we know that if a triangle is a right-angled triangle, then the orthocentre is the vertex of the triangle where the right angle is formed.
Here, orthocentre = vertex B= (-2,1)
(vertex B is already found by solving the equation of line AB and BC).