Asked by natalin
                triangle KLM has vertices K(0,0), L(18,0), and M(6,12).
a. Write equations for the altitudes to the three sides of the triangle.
b. Show that the altitudes intersect at a single point O, called the orthocenter of the triangle.
            
        a. Write equations for the altitudes to the three sides of the triangle.
b. Show that the altitudes intersect at a single point O, called the orthocenter of the triangle.
Answers
                    Answered by
            MathMate
            
    Given:
K(0,0), L(18,0), M(6,12)
To find:
Equations of sides of triangle KLM.
Equations of altitudes of the triangle, namely KK1, LL1, MM1, where K1, L1 and M1 are the intersection of the altitudes of the points K,L,M and the opposite side.
Prove that the three altitudes meet at a single point at the orthocentre.
Equation of a line passing through one point A and perpendicular to a line passing through points B and C is:
(y-ya)(xc-xb)+(x-xa)(yc-yb)=0
Substituting in turn points K, L and M for A,B and C, the equations of the altitudes are:
KK1 : y=x
LL1 : y=9-x/2
MM1 : x=6
The line MM1 is a line parallel to the y-Axis.
Substitute x=6 into the other equations, we get, for the intersection point:
with LL1: y=9-6/2=6, thus (6,6)
with KK1: y=x, thus (6,6)
Therefore the orthocentre is at (6,6).
    
K(0,0), L(18,0), M(6,12)
To find:
Equations of sides of triangle KLM.
Equations of altitudes of the triangle, namely KK1, LL1, MM1, where K1, L1 and M1 are the intersection of the altitudes of the points K,L,M and the opposite side.
Prove that the three altitudes meet at a single point at the orthocentre.
Equation of a line passing through one point A and perpendicular to a line passing through points B and C is:
(y-ya)(xc-xb)+(x-xa)(yc-yb)=0
Substituting in turn points K, L and M for A,B and C, the equations of the altitudes are:
KK1 : y=x
LL1 : y=9-x/2
MM1 : x=6
The line MM1 is a line parallel to the y-Axis.
Substitute x=6 into the other equations, we get, for the intersection point:
with LL1: y=9-6/2=6, thus (6,6)
with KK1: y=x, thus (6,6)
Therefore the orthocentre is at (6,6).
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