Answer:
The orthocentre of the given vertices ( 2 , -3.5)
Step-by-step explanation:
Step(i):-
The orthocentre is the intersecting point for all the altitudes of the triangle.
The point where the altitudes of a triangle meet is known as the orthocentre.
Given Points are K (3.-3), L (2,1), M (4,-3)
The Altitudes are perpendicular line from one side of the triangle to the opposite vertex
The altitudes are MN , KO , LP
step(ii):-
Slope of the line
[tex]KL = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{1-(-3)}{2-3} = -4[/tex]
The slope of MN =
The perpendicular slope of KL
= [tex]\frac{-1}{m} = \frac{-1}{-4} = \frac{1}{4}[/tex]
The equation of the altitude
[tex]y - y_{1} = m( x-x_{1} )[/tex]
[tex]y - (-3) = \frac{1}{4} ( x-4 )[/tex]
4y +12 = x -4
x - 4 y -16 = 0 ...(i)
Step(iii):-
Slope of the line
[tex]LM = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{-3-1}{4-2} = -2[/tex]
The slope of KO =
The perpendicular slope of LM
= [tex]\frac{-1}{m} = \frac{-1}{-2} = \frac{1}{2}[/tex]
The equation of the altitude
[tex]y - y_{1} = m( x-x_{1} )[/tex]
The equation of the line passing through the point K ( 3,-3) and slope
m = 1/2
[tex]y - (-3) = \frac{1}{2} ( x-3 )[/tex]
2y +6 = x -3
x - 2y -9 =0 ....(ii)
Solving equation (i) and (ii) , we get
subtracting equation (i) and (ii) , we get
x - 4y -16 -( x-2y-9) =0
- 2y -7 =0
-2y = 7
y = - 3.5
Substitute y = -3.5 in equation x -4y-16=0
x - 4( -3.5) - 16 =0
x +14-16 =0
x -2 =0
x = 2
The orthocentre of the given vertices ( 2 , -3.5)
