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  • Centroid of a triangle is origin. Circumcentre of the triangle is the fixed point from where all the lines <img alt="\mathrm{a x+b y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cma

Centroid of a triangle is origin. Circumcentre of the triangle is the fixed point from where all the lines \mathrm{a x+b y+c=0} pass, where \mathrm{a, b, c} are distinct non-zero real parameters following the relation \mathrm{a^3+b^3+\mathrm{c}^3=3 \mathrm{abc}.} The orthocentre of the triangle will be
 

Option: 1

\left(-\frac{1}{2}, \frac{1}{2}\right)


 


Option: 2

(-1,-1)
 


Option: 3

(-2,-2)
 


Option: 4

(-2,2)


Answers (1)

best_answer

\mathrm{ \mathrm{a}^3+\mathrm{b}^3+\mathrm{c}^3=3 \mathrm{abc} }

\mathrm{ \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \: \mathrm{as} \: \: \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2 \neq \mathrm{ab}+\mathrm{bc}+\mathrm{ca} \quad(\mathrm{a}, \mathrm{b}, \mathrm{c} \text { distinct }) }
Now \mathrm{ a x+b y+c=0 } is equation of line and on comparing with the relation \mathrm{ x=1, y=1 }. so circumcentre is \mathrm{(1,1)}. Centroid is \mathrm{(0,0),} let orthocentre \mathrm{H\left(x_1, y_1\right)}

\mathrm{0=\frac{x_1+2}{3}, \quad 0=\frac{2+y_1}{3}}

\mathrm{\mathrm{x}_1=-2 \quad \mathrm{y}_1=-2} so orthocentre is (-2,-2)

Hence option 3 is correct.

 

 

Posted by

Pankaj Sanodiya

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