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

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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).