Answer to Question #139719 in Analytic Geometry for Aravind

Question

The coordinates of the vertices of a triangle are A( x1, y1), B( x2, y2), and C( x3, y3)

Then Find the equations of the altitudes of the triangle.and the co ordinates of the point of intersection of altitudes (Orthocentre).

Answer 1

Let ABC be a triangle shown in figure below with ‘O’ as the orthocentre.

Equation of a line = $y - y_{0} = m(x - x_{0})$ where m is the slope of the line.

Equation of line AE = $y - y_{1} = m_{AE}(x - x_{1})$

$m_{AE} * m_{BC} = -1$

$m_{AE} = -1/m_{BC}$

$m_{AE} = (x_{3} - x_{2})/(y_{2} - y_{3})$

Equation of line AE = $y - y_{1} = [ (x_{3} - x_{2})/(y_{2} - y_{3})] * (x - x_{1})$

on simplifying above equation we get

$x(x_{3} - x_2) + y(y_3 - y_2) + x_1(x_2 - x_3) + y_1(y_2 - y_3) = 0$ -----Equation 1

Similarly we will find the equation of line BF

$m_{BF} * m_{AC} = -1$

$m_{BF} = -1/m_{AC}$

$\therefore m_{BF} = (x_3 - x_1)/(y_1-y_ 3)$

Equation of the line BF = $y - y_2 = (x-x_2)*[(x_3-x_1)/(y_1 -y_3)]$

On simplifying above equation we get

$x(x_3-x_1) + y(y_3 - y_1) + x_2(x_1-x_3) + y_2(y_1 -y_3) = 0$ -----equation 2

Equations (1) and (2) show two equations passing through a common point O.Finding the values of x and y,we can get the coordinate of the orthocentre.

Let us find out an easy way out of these variable loophole.

$a_1x+b_1y +c_1 = 0$

$a_1a_2x +a_2b_1y + a_2c_1 = 0$ ---------equation 3

$a_2x +b_2y +c_2 = 0$

$a_2a_1x + a_1b_2y +a_1c_2 = 0$ --------equation 4

Subtracting 3 and 4 and simplifying further we get

$y = (a_1c_2 - c_1a_2)/(a_2b_1 - b_2a_1)$

Similarly we also get :

$a_1x + b_1y +c_1 = 0$

$a_1b_2x +b_2b_1y + c_1b_2 = 0$ -------equation 5

$a_2x+b_2y +c_2 = 0$

$a_2b_1x + b_2b_1y + b_1c_2 = 0$ -------equation 6

On solving we get :

$x = (b_1c_2-c_1b_2)/(a_1b_2-a_2b_1)$

Compare equations (1) and (2) with general form of simultaneous linear equations.

$a_1 = x_3 - x_2$

$a_2 = x_3-x_1$

$b_1 = y_3 - y_2$

$b_2 = y_3 - y_1$

$c_1 = x_1(x_2 - x_3)+ y_1(y_2-y_3)$

$c_2 = x_2(x_1-x_3) + y_2(y_1-y_3)$

substituting the values of $a_1,a_2,b_1,b_2,c_1,c_2$ we get the cordinates of orthocentre O:

$x =( [x_2(x_1-x_3)+y_2(y_1-y_3)][y_3-y_2] -[y_3-y_1][x_1(x_2 - x_3) +y_1(y_2-y_3)]) / [(x_3-x_2)(y_3-y_1) - (y_3-y_2)(x_3 -x_1)]$

$y =( [x_2(x_1-x_3)+y_2(y_1-y_3)][x_3-x_2] -[x_3-x_1][x_1(x_2 - x_3) +y_1(y_2-y_3)]) / [(y_3-y_2)(x_3-x_1) - (x_3-x_2)(y_3 -y_1)]$