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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Aicharya · Accepted Answer

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 $a=-8, b=8$

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,…

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