orthocenter of the triangle with vertices
(1,1),(3,5),(3,0)?
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Given, the vertices of the triangle,
A = (1, 2)
B = (2, 6)
C = (3, -4)
Slope of AB
= y2−y1x2−x1
= 6−22–1
= 41 = 4
Slope of CF
= Perpendicular slope of AB
= −1SlopeofAB
= −14
The equation of CF is given as,
y – y1 = m(x – x1)
y + 4 = −14(x – 3)
4y + 16 = -x + 3
x + 4y = -13 ——————————– (1)
Slope of BC
= y2–y1x2–x1
= −4–63–2
= −101 = -10
Slope of AD
Perpendicular slope of BC
= −1SlopeofBC
= −1−10
= 110
The equation of AD is given as,
y – y1 = m(x – x1)
y + 2 = 110(x – 1)
10y + 20 = x – 1
x – 10y = 21 ——————————– (2)
Subtracting equation (1) and (2),
x + 4y = -13
-x + 10y = -21
——————
14y = -34
y = -2.429
Substituting the value of y in equation (1),
x + 4y = -13
x + 4(-2.429) = -13
x – 9.714 = -13
x = -3.286
Orthocenter = (-3.286,-2.429)
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