Orthocentre of a triangle whose vertices are (0,0),(3,4) and (4,0) is

The orthocenter of a triangle is the point where the three altitudes of the triangle intersect. An altitude of a triangle is a line which passes through a vertex of the triangle and is perpendicular to the opposite side.

Here are the steps to find the orthocenter of the triangle with vertices at (0,0), (3,4), and (4,0):

1. Find the slopes of the lines AB, BC, and AC. The slope of a line through points (x1, y1) and (x2, y2) is (y2 - y1) / (x2 - x1).

   Slope of AB = (4-0) / (3-0) = 4/3 Slope of BC = (0-4) / (4-3) = -4 Slope of AC = (0-0) / (4-0) = 0

2. The slopes of the lines perpendicular to AB, BC, and AC are the negative reciprocals of the slopes of AB, BC, and AC.

   Slope of line perpendicular to AB = -1 / (4/3) = -3/4 Slope of line perpendicular to BC = -1 / -4 = 1/4 Slope of line perpendicular to AC = -1 / 0 = undefined

3. The equations of the lines containing the altitudes of the triangle are y = mx + b, where m is the slope of the line and b is the y-intercept. We can find b by substituting the x and y coordinates of the vertex through which the altitude passes into the equation and solving for b.

   Equation of altitude AD: y = -3/4x + b. Substituting x = 0 and y = 0 gives b = 0, so the equation is y = -3/4x.

   Equation of altitude BE: y = 1/4x + b. Substituting x = 3 and y = 4 gives b = 4 - 3/4 = 7/4, so the equation is y = 1/4x + 7/4.

   The altitude CF is a vertical line through the point (4,0), so its equation is x = 4.

4. The orthocenter of the triangle is the point where the three altitudes intersect. We can find this point by solving the equations of the altitudes for x and y.

   Solving y = -3/4x and y = 1/4x + 7/4 for x gives x = 1.

   Substituting x = 1 into y = -3/4x gives y = -3/4.

   So, the orthocenter of the triangle is at (1, -3/4).