The equation of the line joining the centroid with the orthocentre of the triangle formed by the points $(-2…
The equation of the line joining the centroid with the orthocentre of the triangle formed by the points $(-2,3),(2,-1),(4,0)$ is
$x+y-2=0$
$11 x-y-14=0$
$x-11 y+6=0$
$2 x-y-2=0$
Solution
Let $A D$ and $B E$ are altitudes of the triangle.
$\therefore$ Equation of $A D$ is given by
$
\begin{aligned}
y-3 & =\frac{-1}{\text { Slope of } B C}(x+2) \\
\Rightarrow \quad y-3 & =\frac{-1}{\left(\frac{0+1}{4-2}\right)}(x+2) \\
y-3 & =-2(x+2 \\
y-3 & =-2 x-4 \\
\Rightarrow \quad 2 x+y+1 & =0
\end{aligned}
$
$\therefore$ Equation of $B E$ is given by
$
\begin{aligned}
& y+1=\frac{-1}{\text { Slope of } A C}(x-2) \Rightarrow y+1=\frac{-1}{\left(\frac{0-3}{4+2}\right)}(x-2) \\
& y+1=2(x-2) \\
& y+1=2 x-4 \\
& \Rightarrow \quad y=2 x-5
\end{aligned}
$
Since, orthocentre is the intersecting point of altitudes.
$\therefore$ On solving Eqs. (i) and (ii), we get orthocentre $(1,-3)$.
Also, centroid of $\triangle A B C=\left(\frac{-2+2+4}{3}, \frac{3-1+0}{3}\right)$
$
=\left(\frac{4}{3}, \frac{2}{3}\right)
$
$\therefore$ Equation of line joining $(0,-1)$ and $\left(\frac{4}{3}, \frac{2}{3}\right)$ is
$\begin{aligned}
& y+3=\frac{\frac{2}{3}+3}{\frac{4}{3}-1}(x-1) \\
& \Rightarrow \quad y+3=11(x-1) \\
& \Rightarrow \quad y+3=11 x-11 \\
& \Rightarrow \quad 11 x-y-14=0 . \\
\end{aligned}$