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Geometry 21 Online
OpenStudy (anonymous):

(triangle) ABC has vertices A(0,10),B(4,10),C(-2,4)Find the orthocenter of (triangle) ABC

OpenStudy (anonymous):

@OrangeMaster

OpenStudy (anonymous):

@ChrisTopher987654321

OpenStudy (anonymous):

@jessica_leon93

OpenStudy (anonymous):

@lovelybones69

OpenStudy (anonymous):

@KaTodraC

OpenStudy (anonymous):

don't understand WHAT IT IS ASKING

OpenStudy (anonymous):

Orthocenter is the intersection point of the three lines through a vertex perpendicular to the opposite side. For example, through A, perpendicular to BC; through B, perpendicular to AC; and through C, perpendicular AB. Line through C perpendicular to AB: m = (6-6) / (4-0) = 0 The line is horizontal, so a line perpendicular to that one would be vertical and have an undefined slope. However, since it's going through C, whose x coordinate is 1, the equation for the line through C perpendicular to AB is: x=1 Line perpendicular to BC through A: m of BC = (6-3)/(4-1) = 3/3 = 1 m perpendicular to BC is therefore -1. Line perp. to BC through A ... 6 = -1(0) + b b = 6 Equation of line = y = -1(x) + 6. Line perpendicular to AC through B: m of AC = (6-3)/(0-1) = 3/-1 = -3. m perp to AC = 1/3 Substituting in y=mx+b gives 6 = (1/3)(4) + b b = 14/3 Line equation -> y = (1/3)x + (14/3) Since x must be 1 for all three altitudes, plug in 1 for the other two equations. y = -(x) + 6 y = -(1) + 6 = 5 (1,5) works for the first two equations. Verify in the 3rd. y = (1/3)(1) + (14/3) y = 15/3 = 5. The orthocenter of this triangle is at (1,5).

OpenStudy (anonymous):

ohhh wait I used the wrong numbers, hang on

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