how to find the altitudes of a triangle and the point of intersection of teh altitudes?

Answer 1

The method is entirely dependent on what you know about the triangle in the first place. If you know the 3 sides, you can use Steve's law for the altitude; ex: (alt to side c) = sqrt(4a^2b^2-(a^2+b^2-c^2)^2) / 2c

In general, the altitudes of a triangle intersect at a point called the orthocenter, whose location is described in detail at
http://www.cut-the-knot.org/triangle/altitudes.shtml

Answer 2

the altitude runs from the vertex of one point of a triangle through the opposite side of the triangle at a 90 degree angle
to find the altitude of a triangle, find the slope of one side, take the opposite reciprocal (e.g. 3 to -1/3) and use the points of the opposite vertex to find the equation
the three altitudes in a triangle should all intersect at one point
to find the point, set the equations equal to each other

Answer 3

altitudes of a triangle intersects on the centroid of the triangle enable (h,ok) be the coordinates of centroid h=(x1+x2+x3)/3 & ok=(y1+y2+y3)/3 h=(one million+3+8)/3=4 & ok=(2+8-one million)/3=3 (h,ok)=(4,3) subsequently altitudes meet at (4,3)

Answer 4

knowing what? I assume you are given the 3 sides.

In this case, the area of the triangle, computed by Heron's Formula, is

sqrt(s(s - a)(s - b)(s - c)), where s is the semiperimeter s = (a + b + c)/2.

The area is also 1/2*base*height. Call hA the altitude from side A, hB from B, hC from C. Then hA*A/2 = area = sqrt(s(s - a)(s - b)(s - c))

so hA = 2sqrt(s(s - a)(s - b)(s - c))/A
hB = 2sqrt(s(s - a)(s - b)(s - c))/B
hC = 2sqrt(s(s - a)(s - b)(s - c))/C