What is the orthocenter of a triangle with corners at #(1 ,3 )#, #(6 ,9 )#, and (2 ,4 )#?

Question

What is the orthocenter of a triangle with corners at #(1  ,3   )#, #(6   ,9   )#, and (2    ,4   )#?

Aaliyah Anderson · Accepted Answer

The orthocenter of triangle is #(56,-41)#

Let #triangleABC " be the triangle with corners at"#

#A(1,3), B(6,9) and C(2,4)#

Let #bar(AL) , bar(BM) and bar(CN)# be the altitudes of sides #bar(BC) ,bar(AC) and bar(AB)# respectively.

Let #(x,y)# be the intersection of three altitudes .

Slope of #bar(AB) =(3-9)/(1-6)=6/5#

#bar(AB)_|_bar(CN)=>#slope of # bar(CN)=-5/6# ,

# bar(CN)# passes through #C(2,4)#

#:.#The equn. of #bar(CN)# is #:y-4=-5/6(x-2)#

#6y-24=-5x+10#

#i.e. color(red)(5x+6y=34.....to (1)#

Slope of #bar(BC) =(9-4)/(6-2)=5/4#

#bar(AL)_|_bar(BC)=>#slope of # bar(AL)=-4/5# , # bar(AL)# passes through #A(1,3)#

#:.#The equn. of #bar(AL)# is #:y-3=-4/5(x-1)#

#=>5y-15=-4x+4#

#=>color(red)(4x+5y=19.....to (2)#

Taking #eqn.(1)xx5-eqn.(2)xx(-6)# and adding

#color(white)(....)25x+30y=170#
#ul(-24x-30y=-114#
#:.1x-0=56#

#=>color(blue)( x=56#

From equn.#(2)# we get

#4(56)+5y=19=>5y=19-224=-205=>color(blue)(y=-41#

Hence, the orthocenter of triangle is #(56,-41)#

Allison Allen · Answer

undefined To find the orthocenter of a triangle, we need to first determine the altitudes of the triangle, which are perpendicular lines drawn from each vertex to the opposite side. Then, the point where these altitudes intersect is the orthocenter.

Let's denote the vertices of the triangle as $ A(1, 3) $, $ B(6, 9) $, and $ C(2, 4) $.

We can start by finding the equations of the lines passing through each vertex and perpendicular to the opposite side. The equation of a line passing through a point $ (x_1, y_1) $ with slope $ m $ is given by $ y - y_1 = m(x - x_1) $. For a line perpendicular to a given line with slope $ m $, the slope of the perpendicular line is the negative reciprocal of $ m $.

1. For the line through $ A $ perpendicular to $ BC $ (the line passing through $ B $ and $ C $):
   - Slope of $ BC = \frac{9-4}{6-2} = \frac{5}{4} $
   - Slope of the line perpendicular to $ BC $ passing through $ A $: $ -\frac{4}{5} $
   - Equation of the line: $ y - 3 = -\frac{4}{5}(x - 1) $

2. For the line through $ B $ perpendicular to $ AC $ (the line passing through $ A $ and $ C $):
   - Slope of $ AC = \frac{4-3}{2-1} = 1 $
   - Slope of the line perpendicular to $ AC $ passing through $ B $: $ -1 $
   - Equation of the line: $ y - 9 = -1(x - 6) $

3. For the line through $ C $ perpendicular to $ AB $ (the line passing through $ A $ and $ B $):
   - Slope of $ AB = \frac{9-3}{6-1} = \frac{6}{5} $
   - Slope of the line perpendicular to $ AB $ passing through $ C $: $ -\frac{5}{6} $
   - Equation of the line: $ y - 4 = -\frac{5}{6}(x - 2) $

Next, we find the intersection point of these three lines, which will be the orthocenter of the triangle. Solving these equations simultaneously will give us the coordinates of the orthocenter.