Question

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

• A. x + y - 2 = 0
• B. 11x - y - 14 = 0
• C. x - 11 y + 6 = 0
• D. 2 x - y - 2 = 0

Answer 1

Answer: (B)

Let $AD$ and $BE$ are altitudes of the triangle.
$\therefore$ Equation of $AD$ is given by
$y-3=\frac{-1}{\text{ Slope of }BC}(x+2)$
$\Rightarrow y-3=\frac{-1}{\left(\frac{0+1}{4-2}\right)}(x+2)$
$y-3=-2(x+2)$
$y-3=-2x-4$
$\Rightarrow 2x+y+1=0$.......(i)
$\therefore$ Equation of $BE$ is given by
$y+1=\frac{-1}{\text{ Slope of }AC}(x-2)$
$\Rightarrow y+1=\frac{-1}{\left(\frac{0-3}{4+2}\right)}(x-2)$
$y+1=2(x-2)$
$y+1=2x-4$
$\Rightarrow y=2x-5$.......(ii)
Since, orthocentre is the intersecting point of altitudes.
$\therefore$ On solving Eqs. (i) and (ii), we get orthocentre $(1,-3)$
Also, centroid of $△ABC=\left(\frac{-2+2+4}{3},\frac{3-1+0}{3}\right)$
$=\left(\frac{4}{3},\frac{2}{3}\right)$
$\therefore$ Equation of line joining $(0,-1)$ and $\left(\frac{4}{3},\frac{2}{3}\right)$ is
$y+3=\frac{\frac{2}{3}+3}{\frac{4}{3}-1}(x-1)$
$\Rightarrow y+3=11(x-1)$
$\Rightarrow y+3=11x-11$
$\Rightarrow 11x-y-14=0$