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How To Find Orthocenter Algebraically

The orthocenter is the intersecting bespeak for all the altitudes of the triangle. The signal where the altitudes of a triangle come across is known as the Orthocenter. In the beneath mentioned diagram orthocenter is denoted by the letter 'O'. There is no direct formula to calculate the orthocenter of the triangle. It lies inside for an acute and outside for an obtuse triangle.

Altitudes are null but the perpendicular line ( Advert, Exist and CF ) from ane side of the triangle ( either AB or BC or CA ) to the opposite vertex. Vertex is a point where 2 line segments meet ( A, B and C ).

Orthocenter of a TriangleTo Summate the slope of the sides of the triangle. The formula to calculate the gradient is given as,

\[\big Slope\;of\;a\;Line=\frac{y_{2}-y_{one}}{x_{two}-x_{ane}}\]

To calculate the perpendicular slope of the sides of the triangle. It gives u.s.a. the slope of the altitudes of the triangle. The formula to summate the perpendicular gradient is given every bit,

\[\large Perpendicular\;Slope\;of\;a\;Line=\frac{-1}{Slope\;of\;a\;Line}\]

To calculate the equation for the altitudes with their respective coordinates. The point-slope formula is given as,

\[\big y-y_{1}=m(x-x_{1})\]

Finally, by solving any two altitude equations, nosotros can get the orthocenter of the triangle.

Solved Example

Question: Find the orthocenter of a triangle when their vertices are A(1, two), B(2, 6), C(3, -4).

Solution:

Given, the vertices of the triangle,

A = (one, two)
B = (2, 6)
C = (3, -4)

Slope of AB

= \(\frac{y_{two}-y_{1}}{x_{2}-x_{i}}\)

= \(\frac{vi − 2}{ii – i}\)

= \(\frac{4}{1}\)= 4

Slope of CF

= Perpendicular slope of AB

= \(\frac{-i}{Slope of AB}\)

= \(\frac{-1}{4}\)

The equation of CF is given as,

y – yone = m(x – tenane)

y + 4 = \(\frac{-1}{4}\)(10 – iii)

4y + sixteen = -x + iii

x + 4y = -xiii ——————————– (1)

Slope of BC

=\(\frac{y_{2} – y_{1}}{x_{two} – x_{one}}\)

= \(\frac{-4 – 6}{3 – two}\)

= \(\frac{-10}{1}\)= -10

Slope of AD

Perpendicular slope of BC

= \(\frac{-1}{Gradient of BC}\)

= \(\frac{-1}{-10}\)

= \(\frac{one}{x}\)

The equation of Advertising is given as,

y – y1 = m(x – x1)

y – 2 = 1/x(10 – 1)

10y – 20 = x – i

x – 10y = -19 ——————————– (2)

Subtracting equation (2) and (1),

x  + 4y  = -xiii

-ten  + 10y =  19

——————

14y = 6

y = 6/14 = iii/7

Or

y = 0.429

Substituting the value of y in equation (1),

x + 4y = -xiii

10 + 4(0.429) = -13

ten + 1.716 = -13

x = -14.716

Orthocenter = (-14.716, 0.429)

Source: https://byjus.com/orthocenter-formula/

Posted by: kahnpritter.blogspot.com

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