Find the orthocenter of the triangle with the given vertices: X(-5, 4), Y(2, -3), Z(1, 4).

To find the orthocentre  of XYZ, X(-5,4), Y(2,-3),
Z(1,4).

Orthocentre O is the common intersecting points of
altitudes through X, Y, and Z at O.

Since , the equation of
the altitude though Z  (1 ,4) is perpendicular to XY, its slope = - 1/{(-3 -4)/(2- -5))}
= 7/7 = 1. So the equation of the altitude  with slope = 1 through Z(1,4) has the
equation: y-4 = 1(x-1) .

Or y -x =
3....(1).

The slope of the altitude through X (-5,4) is 
-1/{(4- -3)/(1-2)} = 1/7. Therefore the equation of the altitude with slope 1/7 and
through X(-5,4) is y - 4 = (1/7)(x- -5).  Or  y- x/7 = 5/7
+4.

Or y - x/7 =
33/7........(2).

Now we solve for the intersection point of
the altitudes at (1) and (2) to get the coordinates of the
orthocentre:

EQq(1) - (2) gives:  -x+x/7 = 3- 33/7 =
-12/7

-6x/7 = -12/7.

x =
-12/-6 = 2.

So x= 2.

Therefore
from (1) , y -x = 3. Therefore y = x+3 = 2+3 = 5.

Therefore
the coordinates of the orthocentre of XYZ = (2, 5).