Answer:
The coordinates of orthocenter of the triangle are (1,5).
Step-by-step explanation:
The given vertices are A(0,6), B(4,6) and C(1,3).
Orthocenter of a triangle is the intersection point of all altitudes.
The product of slopes of two perpendicular lines is -1.
The slope of AB is
[tex]m_{AB}=\frac{y_2-y_1}{x_2-x_1}=\frac{6-6}{4-0}=0[/tex]
The slope of AB is 0, therefore slope of line which is perpendicular to AB is [tex]\frac{-1}{0}[/tex].
Point slope form of a line is
[tex]y-y_1=m(x-x_1)[/tex]
Where, m is the slope.
The equation of altitude on AB form C is
[tex]y-3=\frac{1}{0}(x-1)[/tex]
[tex]0=x-1[/tex]
[tex]x=1[/tex] ..... (1)
Slope of BC is
[tex]m_{BC}=\frac{y_2-y_1}{x_2-x_1}=\frac{6-3}{1-4}=1[/tex]
The slope of BC is 1 therefore the slope of altitude on BC from A is -1.
The equation of altitude on BC from A is
[tex]y-6=-1(x-0)[/tex]
[tex]y=-x+6[/tex] ..... (2)
Using (1) and (2) we get
[tex]x=1, y=5[/tex]
The intersection point of two altitudes is (1,5). Therefore coordinates of orthocenter of the triangle are (1,5).
