What is the orthocenter of a triangle with corners at #(6 ,3 )#, #(2 ,4 )#, and (7 ,9 )#?
Orthocenter of the triangle is at
Orthocenter is the point where the three "altitudes" of a triangle meet. An "altitude" is a line that goes through a vertex (corner point) and is at right angles to the opposite side.
Solving equation(1) and (2) we get their intersection point , which
is the orthocenter. Adding equation(1) and (2) we get,
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To find the orthocenter of a triangle formed by the points (6,3), (2,4), and (7,9), we first need to find the equations of the altitudes of the triangle. An altitude of a triangle is a perpendicular line from one vertex to the opposite side (or the line extended). The intersection point of the altitudes is the orthocenter.
Let's consider the vertices A(6,3), B(2,4), and C(7,9).

Find the slope of BC: The slope of a line through points (x1, y1) and (x2, y2) is given by (y2y1) / (x2x1). So, the slope of BC is (94) / (72) = 5/5 = 1.

Slope of the altitude from A: The slope of the altitude from A to BC will be the negative reciprocal of the slope of BC, since it's perpendicular. So, the slope of the altitude from A is 1.

Equation of the altitude from A: Using the pointslope form of a line equation, y  y1 = m(x  x1), where m is the slope and (x1, y1) is a point on the line. With slope 1 and passing through A(6,3), the equation is y  3 = 1(x  6), which simplifies to y = x + 9.
Repeat this process for another pair to find the second altitude:

Find the slope of AC: The slope is (93) / (76) = 6/1 = 6.

Slope of the altitude from B: The negative reciprocal of 6 is 1/6.

Equation of the altitude from B: With slope 1/6 and passing through B(2,4), the equation is y  4 = 1/6(x  2), simplifying to y = (1/6)x + (25/6).
Finally, find the intersection of these two altitudes:
Solving the equations y = x + 9 and y = (1/6)x + (25/6) simultaneously will give us the orthocenter's coordinates.
Setting x + 9 = (1/6)x + (25/6):

Multiply everything by 6 to clear the fraction: 6x + 54 = x + 25

Bring terms involving x to one side: 6x + x = 25  54

Simplify: 5x = 29

Solve for x: x = 29/5 = 5.8
Substitute x = 5.8 back into one of the line equations to find y:
y = 5.8 + 9 = 3.2
Therefore, the orthocenter of the triangle is at the point (5.8, 3.2).
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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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