If the sides of a triangle ABC are 2x²-y²=0,x+y-1=0 and the sides of another triangle PQR are 2x²-5xy+2y²=0, 7x-2y-12=0, then the distance between the centroid of ΔABC and the orthocenter of ΔPQR is

Correct Answer is : 131651/3{165}31165

Correct Answer is : 131651/3{165}31165

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TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 48 of 160

Marks: +1, -0

2x^2-y^2=0,x+y-1=0

2x^2-5xy+2y^2=0, 7x-2y-12=0,

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

\Delta 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}

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