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If the straight line $2 x+3 y-1=0, x+2 y-1=0$ and $a x+b y-1=0$ form a triangle with origin as orthocentre, then $(a, b)$ is equal to
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Verified Answer
The correct answer is:
$(-8,8)$
Here, point $A$ is the intersection of line $A B$ and $A C$ so equation of line passing through $A$.
This line passes through the orthocentre $(0,0)$, then
$\begin{aligned}
& -1-\lambda=0 \\
& \lambda=-1
\end{aligned}$
On substituting $\lambda=-1$ in Eq. (i), we get $x+y=0$ as the equation of $A D$. Since $A D \perp B C$, therefore
$-1 \times-\frac{a}{b}=-1$
Similarly, by applying the condition that $B E$ is perpendicular to $C A$, we get
Now, solving Eqs. (ii) and (iii), we get $a=-8, b=8$
This line passes through the orthocentre $(0,0)$, then
$\begin{aligned}
& -1-\lambda=0 \\
& \lambda=-1
\end{aligned}$
On substituting $\lambda=-1$ in Eq. (i), we get $x+y=0$ as the equation of $A D$. Since $A D \perp B C$, therefore
$-1 \times-\frac{a}{b}=-1$
Similarly, by applying the condition that $B E$ is perpendicular to $C A$, we get
Now, solving Eqs. (ii) and (iii), we get $a=-8, b=8$
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