Answer:
The orthocentre of the given vertices ( 2 , -3.5)
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
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)
<em>The Altitudes are perpendicular line from one side of the triangle to the opposite vertex</em>
<em>The altitudes are MN , KO , LP</em>
<u><em>step(ii):-</em></u>
<em> </em> Slope of the line
![KL = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{1-(-3)}{2-3} = -4](https://tex.z-dn.net/?f=KL%20%3D%20%5Cfrac%7By_%7B2%7D-y_%7B1%7D%20%20%7D%7Bx_%7B2%7D-x_%7B1%7D%20%20%7D%20%3D%20%5Cfrac%7B1-%28-3%29%7D%7B2-3%7D%20%3D%20-4)
The slope of MN =
The perpendicular slope of KL
= ![\frac{-1}{m} = \frac{-1}{-4} = \frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7Bm%7D%20%3D%20%5Cfrac%7B-1%7D%7B-4%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D)
The equation of the altitude
![y - y_{1} = m( x-x_{1} )](https://tex.z-dn.net/?f=y%20-%20y_%7B1%7D%20%3D%20m%28%20x-x_%7B1%7D%20%29)
![y - (-3) = \frac{1}{4} ( x-4 )](https://tex.z-dn.net/?f=y%20-%20%28-3%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%28%20x-4%20%29)
4y +12 = x -4
x - 4 y -16 = 0 ...(i)
Step(iii):-
Slope of the line
![LM = \frac{y_{2}-y_{1} }{x_{2}-x_{1} } = \frac{-3-1}{4-2} = -2](https://tex.z-dn.net/?f=LM%20%3D%20%5Cfrac%7By_%7B2%7D-y_%7B1%7D%20%20%7D%7Bx_%7B2%7D-x_%7B1%7D%20%20%7D%20%3D%20%5Cfrac%7B-3-1%7D%7B4-2%7D%20%3D%20-2)
The slope of KO =
The perpendicular slope of LM
= ![\frac{-1}{m} = \frac{-1}{-2} = \frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7Bm%7D%20%3D%20%5Cfrac%7B-1%7D%7B-2%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
The equation of the altitude
![y - y_{1} = m( x-x_{1} )](https://tex.z-dn.net/?f=y%20-%20y_%7B1%7D%20%3D%20m%28%20x-x_%7B1%7D%20%29)
The equation of the line passing through the point K ( 3,-3) and slope
m = 1/2
![y - (-3) = \frac{1}{2} ( x-3 )](https://tex.z-dn.net/?f=y%20-%20%28-3%29%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28%20x-3%20%29)
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)