Question

Assertion :Affix of the orthocentre of $△ABC$ is $z_1+z_2+z_3-2z_0$. Reason: $z_4$ is given by $z_0-\frac{(z_0-z_2)(z_0-z_3)}{z_0-z_1}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Centroid $G$ of $△ABC$ is $\frac{1}{3}(z_1+z_2+z_3)$
Let orthocenter $H$ of $△ABC$ be $z.$
As $G$ divides the join of $H$ and $S$ in the ratio $2:1,$ we get
$\frac{1}{3}(z_1+z_2+z_3)=\frac{{2z}_0+z}{2+1}\Rightarrow z=z_1+z_2+z_3-{2z}_0$
As $AD\perp BC,\frac{z_4-z_1}{z_2-z_3}$ is purely imaginary
$\Rightarrow \frac{z_4-z_1}{z_2-z_3}+\frac{\overline{z}-\overline{z_1}}{\overline{z_2}-\overline{z_3}}=0\Rightarrow \frac{w_4-w_1}{w_2-w_3}+\frac{\overline{w}-\overline{w_1}}{\overline{w_2}-\overline{w_3}}=0$
where $w_i=z_i-z_0$ for $i=1,2,34$
As $z_1,z_2,z_3,z_4$ lie on a circle with center at $z_0$
$|z_i-z_0|=r\Rightarrow w_i\overline{w_i}=r^2$ for $i=1,2,3,4$
Now $\frac{w-w_1}{w_2-w_3}+\frac{\frac{r^2}{w_4}-\frac{r^2}{w_i}}{\frac{r^2}{w_2}-\frac{r^2}{w_3}}=0\Rightarrow \frac{w-w_1}{w_2-w_3}[1+\frac{w_2{w}_3}{{ww}_1}]=0$
$\Rightarrow w=-\frac{w_2{w}_3}{w_1}$