How to find the coordinates of intersection of altitudes in a triangle

Consider the triangle with vertices A, B, C, the coordinates of which are respectively (xa, ya), (xb, yb), (xc, yc). Guide the height of the vertices of the triangle and label the point of intersection of altitudes as the point with coordinates (x, y) and must find.

Write down the equation of the sides of the triangle. Side AB is expressed by the equation (x−xa)/(xb−xa)=(y−ya)/(yb−ya). Give the equation to the form y=k×x+b: x×yb of x×ya−xa×yb+xa×ya=y×xb−y×xa−ya×xb+ya×xa, which is equivalent to y=((yb−ya)/(xb−xa))×x+xa×(ya−yb)/(xb−xa)+ya. Denote the angular coefficient of k1=(yb−ya)/(xb−xa). Similarly find the equation of any other side of the triangle. Side AC is given by the formula (x−xc)/(xa−xc)=(y−yc)/(ya−yc), y=((ya−yc)/(xa−xc))×x+xc×(ya−yc)/(xc−xa)+ya. The angular coefficient of k2=(yc−yb)/(xc−xb).

Write down the equations of the triangle's altitudes drawn from vertices B and C. since the height of the emerging from vertex B perpendicular to side AC, then its equation would be y−ya=(-1/k2)×(x−xa). And a height extending perpendicular to the side AB, and emerging from the point C, will be expressed in the form y−yc=(-1/k1)×(x−xc).

Find the point of intersection of two altitudes of a triangle, solving a system of two equations with two unknowns: y−ya=(-1/k2)×(x−xa) and y−yb=(-1/k1)×(x−xb). Express the variable y from both the equations, Paranaita these expressions and solve the equation with respect to x. And then substitute the resulting value of x into one of the equations and find y.

Let's consider for a better understanding of the issue example. Suppose that we are given a triangle with vertices A (-3, 3), B (5, -1) and C (5, 5). Write down the equation of the sides of the triangle. Side AB is expressed by the formula (x+3)/(5+3)=(y−3)/(-1−3) or y=(-1/2)×x+3/2, that is, k1=-1/2. Side AC is given by equation (x+3)/(5+3)=(y−3)/(5-3), that is y=(1/4)×x+15/4. The angular coefficient k2=1/4. The equation of the altitude coming out of vertex C is: y−5=2×(x−5) or y=2×x−5, and the height coming from the top of B: y−5=-4×(x+1), that is y=-4×x+19. Solve the system of these two equations. It turns out that the orthocenter has coordinates (4, 3).