Answer:
The centroid of MNP is [tex](0,2)[/tex] and the orthocenter of the MNP is [tex](-2,\frac{-2}{3})[/tex].
Step-by-step explanation:
The vertices of triangle are M (-4, -2), N (6, -2), and P (-2, 10).
Centroid of a triangle is the intersection point of all medians.
Formula for centroid
[tex](\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})[/tex]
The centroid of MNP is
[tex](\frac{-4+6-2}{3},\frac{-2-2+10}{3})[/tex]
[tex](\frac{0}{3},\frac{6}{3})[/tex]
[tex](0,2)[/tex]
Orthocenter of a triangle is the intersection point of all altitudes.
We have to find the intersection of at least two altitudes.
Slope of a line is
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
From the figure it is noticed that the line MN is a horizontal line, slope of MN is
[tex]m=\frac{-2+2}{6+4}=0[/tex]
The product of slopes of perpendicular lines is -1. Therefore the product of slopes of MN and AP is -1.
[tex]m_{MN}\times (m_{AP})=-1[/tex]
[tex]0\times (m_{AP})=-1[/tex]
[tex]m_{AP}=\frac{-1}{0}[/tex]
Point slope form
[tex]y-y_1=m(x-x_1)[/tex]
Where m is slope.
Equation of AP is
[tex]y-10=\frac{-1}{0}(x+2)[/tex]
[tex]0=x+2[/tex]
[tex]x=-2[/tex] ..... (1).
Slope of MP is
[tex]m_{MP}=\frac{10+2}{-2+4}=\frac{12}{2}=6[/tex]
The product of slopes of perpendicular lines is -1. Therefore the product of slopes of MP and NB is -1.
[tex]m_{MP}\times (m_{NB})=-1[/tex]
[tex]6\times (m_{NB})=-1[/tex]
[tex]m_{NB}=\frac{-1}{6}[/tex]
Equation of NB is
[tex]y+2=\frac{-1}{6}(x-6)[/tex]
[tex]6y+12=-x+6[/tex]
[tex]x+6y=-6[/tex] ..... (2)
Using equation (1) we get
[tex](-2)+6y=-6[/tex]
[tex]6y=-4[/tex]
[tex]y=-\frac{2}{3}[/tex]
Therefore the orthocenter of the triangle is [tex](-2,\frac{-2}{3})[/tex].
