If two vertices of a triangle are $\left(-2,3\right)$ and $\left(5,-1\right)$ , orthocenter lies at the origin and centroid on the line $x+y=7$ , then the third vertex lies at

Given O (0, 0) is the orthocenter.

Let A (h, k) be third vertex,$B\left(-2,3\right)$ , and $C\left(5,-1\right)$ the other two vertices.

Then the slope of the line through A and O is k/ h, while the lines through B and C have the slope$-\frac{4}{7}$ . By the property of the orthocenter, these  two lines must be perpendicular, so we have

 $\left(\frac{k}{h}\right)(-\frac{4}{7})=-1\Rightarrow \frac{k}{h}=\frac{7}{4}\to \left(1\right)$

Also, $\frac{5-2+h}{3}+\frac{-1+3+k}{3}=7$

$\Rightarrow h+k=\mathrm{16.........}\left(2\right)$

Which is not satisfied by the points given in (a), (b), or (c)