Two vertices of a triangle are (4, -3) and (-2, 5). If the orthocenter of the triangle is at (1,2),then the third vertex is

Let the third vertex be (h, k).

Now, the slope of AD is (k - 2)/(h - 1), the slope of BC is (5 + 3)l(-2 - 4)=4/3, the slope of BE is (-3 -2)/(4 - 1) = 5/3, and the 

slope of AC is (k - 5)/(h + 2).

Since  $\mathrm{AD}\perp \mathrm{BC},$
we have $\frac{\mathrm{k}-2}{\mathrm{h}-1}\times \frac{-4}{3}=-1$
or 3h-4k+5=0 (i)
Again, since $\mathrm{BE}\perp \mathrm{AC}$, we have
$-\frac{5}{3}\times \frac{\mathrm{k}-5}{\mathrm{h}+2}=-1$
or 3h - 5k +31 = 0 (ii)
On solving (i) and (ii), we get h=33, k=26.Hence, the third vertex is (33, 26).