Question

The altitude from the vertices $A,B$ and $C$ of the triangle $ABC$ meet its circumcircle at $D,E$ and $F$ respectively. The complex numbers representing the points $D,E$ and $F$ $\text{are}{z}_1,z_2\text{and}{z}_3\text{respectively. If}\frac{z_3-z_1}{z_2-z_1}\text{is purely}$ real, then the triangle $ABC$ is right angle at

• A. $B$
• B. $A$
• C. $C$
• D. None of these

Answer 1

The correct option is B $A$


$\angle FDA=\angle FCA={90}^{\circ }-A$
$\angle ADE=\angle ABE={90}^{\circ }-A$
$\Rightarrow \angle FDE={180}^{\circ }-2A=\pi -2A$
Similarly, $\angle DFE=\pi -2C\text{and}\angle DEF=\pi -2B$
The angles of $\Delta DEF$ are $\pi -2A,\pi -2B\text{and}\pi -2C$, respectively.

$\text{Also, it is given that}\frac{z_3-z_1}{z_2-z_1}\text{is purely real.}$
$\text{Hence,}\arg \left(\frac{z_3-z_1}{z_2-z_1}\right)=0\text{or}\pi$
$\Rightarrow \pi -2A=0\text{or}\pi$
$\Rightarrow A=\frac{\pi }{2}\text{or}0\left(\text{not permissible}\right)$
Hence, triangle $ABC$ is right angled at $A.$