Instruction

1

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.2

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).

3

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).

4

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.5

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).