What is the orthocenter of a triangle with corners at #(3 ,1 )#, #(1 ,3 )#, and (5 ,2 )#?
The orthocenter is at the point
Any perpendicular line's slope, n, is
To get the desired line equation, use the point slope form of the equation of a line:
Any perpendicular line has a slope, n, which is:
The point where these two lines intersect is the orthocenter:
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To find the orthocenter of a triangle with vertices (3,1), (1,3), and (5,2), follow these steps:
 Determine the slopes of the lines passing through each pair of vertices.
 Find the perpendicular bisectors of the sides of the triangle.
 The point of intersection of the perpendicular bisectors is the orthocenter.
Let's find the slopes of the sides of the triangle:

Slope of the line passing through (3,1) and (1,3): ( m_1 = \frac{{3  1}}{{1  3}} = 1 )

Slope of the line passing through (3,1) and (5,2): ( m_2 = \frac{{2  1}}{{5  3}} = \frac{1}{2} )

Slope of the line passing through (1,3) and (5,2): ( m_3 = \frac{{2  3}}{{5  1}} = \frac{1}{4} )
Now, find the equations of the perpendicular bisectors:

For the line passing through (3,1) and (1,3), the midpoint is ( \left(\frac{{3 + 1}}{2}, \frac{{1 + 3}}{2}\right) = (2,2) ). Its slope will be the negative reciprocal of ( m_1 ), so it is ( m_1' = 1 ). The equation of the perpendicular bisector is ( y  2 = 1(x  2) ) or ( y = x ).

For the line passing through (3,1) and (5,2), the midpoint is ( \left(\frac{{3 + 5}}{2}, \frac{{1 + 2}}{2}\right) = (4,1.5) ). Its slope will be the negative reciprocal of ( m_2 ), so it is ( m_2' = 2 ). The equation of the perpendicular bisector is ( y  1.5 = 2(x  4) ) or ( y = 2x + 9 ).

For the line passing through (1,3) and (5,2), the midpoint is ( \left(\frac{{1 + 5}}{2}, \frac{{3 + 2}}{2}\right) = (3,2.5) ). Its slope will be the negative reciprocal of ( m_3 ), so it is ( m_3' = 4 ). The equation of the perpendicular bisector is ( y  2.5 = 4(x  3) ) or ( y = 4x  10 ).
Solve the system of equations to find the point of intersection, which is the orthocenter.
( y = x ) (1)
( y = 2x + 9 ) (2)
( y = 4x  10 ) (3)
Solving (1) and (2), we get ( x = 2 ) and ( y = 2 ).
Solving (1) and (3), we get ( x = \frac{5}{3} ) and ( y = \frac{5}{3} ).
Therefore, the orthocenter of the triangle is ( \left(\frac{5}{3}, \frac{5}{3}\right) ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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