What is the orthocenter of a triangle with corners at #(7 ,3 )#, #(4 ,8 )#, and (6 ,3 )#?

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Answer 1 · 2016-11-07T18:04:52

The orthocenter is #(4, 9/5)#

Determine the equation of the altitude that goes through point #(4,8)# and intersects the line between the points #(7,3) and (6,3)#.

Please notice that the slope of line is 0, therefore, the altitude will be a vertical line:

#x = 4##" [1]"#

This is an unusual situation where the equation of one of the altitudes gives us the x coordinate of the orthocenter, #x = 4#

Determine the equation of the altitude that goes through point #(7,3)# and intersects the line between the points #(4,8) and (6,3)#.

The slope, m, of the line between the points #(4,8) and (6,3)# is:

#m = (3 - 8)/(6 - 4) = -5/2#

The slope, n, of the altitudes will be the slope of a perpendicular line:

#n = -1/m#

#n = 2/5#

Use the slope, #2/5#, and the point #(7,3)# to determine the value of b in the slope-intercept form of the equation of a line, #y = nx + b#

#3 = (2/5)7 + b#

#b = 3 - 14/5#

#b = 1/5#

The equation of the altitude through point #(7,3)# is:

#y = (2/5)x + 1/5##" [2]"#

Substitute the x value from equation [1] into equation [2] to find the y coordinate of the orthocenter:

#y = (2/5)4 + 1/5#

#y = 9/5#

The orthocenter is #(4, 9/5)#