Q88 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

If the sides of a triangle $A B C$ are $2 x^2-y^2=0$, $x+y-1=0$ and the sides of another triangle $P Q R$ are $2 x^2-5 x y+2 y^2=0,7 x-2 y-12=0$, then the distance between the centroid of $\triangle A B C$ and the orthocentre of $\triangle P Q R$ is

  • A
    $\frac{4}{3} \sqrt{261}$
  • B
    $\frac{1}{3} \sqrt{165}$
  • Correct answer
    C
    $2 \sqrt{29}$
  • D
    $56 \sqrt{3}$

Solution

For $\triangle A B C$, we have sides are $2 x^2-y^2=0, x+y-1=0$ $(\sqrt{2} x-y)(\sqrt{2} x+y)=0, x+y-1=0$ $\sqrt{2} x-y=0, \sqrt{2} x+y=0, x+y-1=0$ On solving these, we get vertices $(0,0),(\sqrt{2}-1,2-\sqrt{2}),(-\sqrt{2}-1,2+\sqrt{2})$ For $\triangle P Q R$, we have sides are $2 x^2-5 x y+2 y^2=0,7 x-2 y-12=0$ $(2 x-y)(x-2 y)=0,7 x-2 y-12=0$ $2 x-y=0, x-2 y=0,7 x-2 y-12=0$ On solving these, we get vertices $(0,0),(4,8),(2,1)$ Now, centroid of $\triangle A B C$ $G=\left(\frac{0+\sqrt{2}-1-\sqrt{2}-1}{3}, \frac{0+2-\sqrt{2}+2+\sqrt{2}}{3}\right)$ $=\left(\frac{-2}{3}, \frac{4}{3}\right)$ Orthocentre of $\triangle P Q R$, $H=\left(\frac{(8-1)(8+8)}{16-4}, \frac{(2-4)(8+8)}{16-4}\right)=\left(\frac{28}{3},-\frac{8}{3}\right)$ $\therefore$ Required distance=G H $=\sqrt{\left(\frac{28}{3}+\frac{2}{3}\right)^2+\left(\frac{-8}{3}-\frac{4}{3}\right)^2}$ $=\sqrt{100+16}=\sqrt{116}=2 \sqrt{29}$

Hence, the correct answer is option: 'C'.

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