Question

If the straight lines $2x+3y-1=0$, $x+2y-1=0$ and $ax+by-1=0$ from a triangle with the origin as orthocenter, then $(a,b)$ is givne by

• A. $(6,4)$
• B. $(-3,3)$
• C. $(8,-8)$
• D. $(0,7)$

Answer 1

Answer: (C)

$(8,-8)$

On a $△ABC$ $AB:2x+3y-1=0$ $AC:x+2y-1=0$

O be the point of intersection of the perpendiculars

$\begin{array}{cc}AD:& 2x+3y-1+\lambda (x+2y-1)=0\\ \Rightarrow & x(2+\lambda )+y(3+2\lambda )+(-1-\lambda )=0\\ \Rightarrow & 0(2+\lambda )+0(3+2\lambda )+(-1-\lambda )=0\\ -1-\lambda & =0\\ \lambda & =-1\\ & x(2-1)+y(3-2)+0=0\\ & A0:x+y=0\end{array}$

$\begin{array}{cc}& AD\text{is}\perp \text{to}BC\\ & \begin{array}{c}{m}_{A0}\times m_{BC}=-1\\ (-1)(-\frac{a}{b})=-1\\ a=-b\\ B0:(2x+3y-1)+\mu (ax+by-1)=0\\ x(2+a\mu )+y(3-a\mu )+(-1-\mu )=0\\ \mu =-1\\ \Rightarrow x(2-a)+y(3+a)=0\\ m_{B0}=-\left(\frac{2-a}{3+a}\right)\end{array}\end{array}$

$\begin{array}{cc}BO& 19\text{Lto}AB\\ \Rightarrow & m_{BO}\times m_{AC}=-1\end{array}$

$\begin{array}{cc}-\left(\frac{2-a}{3+a}\right)\times (-\frac{1}{2})& =-1\\ \frac{2-a}{3+a}=-2& \Rightarrow 2-a=-6-2a\\ & \Rightarrow a=-8\\ & \text{and}b=8\\ & (a,b)=(8,-8)\\ \text{Hence,}\left(C\right)& \text{is the correct option}\end{array}$