Suppose a triangle of area 27 sq. units is formed by $18 x^2-9 x y+y^2=0$ and $y=c$. Then the centroid of…
Suppose a triangle of area 27 sq. units is formed by $18 x^2-9 x y+y^2=0$ and $y=c$. Then the centroid of the triangle is
$(3,12)$
$(12,3)$
$(-12,3)$
$(-3,12)$
Solution
Let two lines $4=\mathrm{y}=\mathrm{m}, \mathrm{x}$ and $\mathrm{b}=\mathrm{y} \mathrm{m}$
$\Rightarrow 18 x 2-9 x y+y 2=(y-m, x)(y-m, x)$
on complaining $\mathrm{M}_1+\mathrm{M}_2=\mathrm{g}$ and $\mathrm{M}_1, \mathrm{M}_2=18$
$\Rightarrow M_1=3,6 \quad M_2=6,3$
hence equations are $y=3 x, y=6 x$ now point of inter section of $y=3 x, y=6 x$ and $y=c$ will be
$(0,0)\left(\frac{\mathrm{c}}{6}, \mathrm{c}\right)$ and $\left(\frac{\mathrm{c}}{3}, \mathrm{c}\right)$
now area of $\varnothing \Rightarrow \frac{1}{2} \times \frac{\mathrm{c}}{6} \times \mathrm{c}=27$ (Given)
$\Rightarrow \mathrm{C}=18$
Co ordinates will be $(0,0),(3,18)$ and $(6,18)$ now Centroid $=(3,12)$