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If the straight lines $\mathrm{2 x+3 y=1, x+2 y=1,}$ and $\mathrm{a x+b y=1}$ form a triangle with origin as orthocenter, then $\mathrm{(a, b)}$ is

Option: 1
$(6,4)$

Option: 2
$(4,6)$

Option: 3
$(-8,8)$

Option: 4
$(8,-8)$

Answers (1)

best_answer

Equation of $\mathrm{\mathrm{AO}\: is \: 2 \mathrm{x}+3 \mathrm{y}-1+\lambda(\mathrm{x}+2 \mathrm{y}-1)=0 \:}$

$\mathrm{ where \: \lambda=-1}$

$\mathrm{\mathrm{AO} \perp \mathrm{BC} }$

$\mathrm{\therefore }$ line passing through origin $\mathrm{\mathrm{x}+\mathrm{y}=0 }$

$\mathrm{-1\times \frac{a}{-b}=-1 }$

$\mathrm{ \Rightarrow \mathrm{a}=-\mathrm{b} }$

Equation $\mathrm{ \mathrm{BO}\: is \: 2 \mathrm{x}+3 \mathrm{y}-1+\mu(\mathrm{ax}+\mathrm{by}-1)=0 }$

$\mathrm{ \mu=-1 }$

$\mathrm{ \mathrm{BO} \perp \mathrm{AC} }$

Now equation of $\mathrm{ B O\: is \: (2-a) x+(3-b) y=0}$

$\mathrm{ \frac{-(2-a)}{3+a} \times-\frac{1}{2}=-1 \Rightarrow a=8, b=-8 }$

Hence option 3 is correct.

Posted by

Deependra Verma

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