Task:

Find the equations of the sides of a triangle, given one its vertex A(3;-1), and also the equation of the bisector x-4y+10=0 and the median 6x+10y-59=0 drawn from different vertices.

Solution:

Let's find coordinates of point C.

Since BD is a median, coordinates of point D can be found as a coordinates of the middle of segment.

Let's substitute the obtained points into the equation of the median 6x+10y-59=0:

6()+10()-59=0

3х_с+5у_с-55=0

Since AC∩CL at point C, we find the coordinates of point C from the solution of the system

Now we have point C(10;5).

Let's make the equation of the line AC passing through point A(3;-1) and point C(10;5) by the formula:

Side AC: 6х-7у-25=0

Let's make the equation of line BC, which is inclined to line AC at the angle 2γ.

Since CL is the bisector of angle C (angle 2γ), we find tg γ at intersecting lines AC and CL

АС: 6х-7у-25=0; у=6/7х – 25/7, k_AC=6/7

CL: х-4у+10=0; у=1/4х + 10/4, k_CL=1/4

By the formula for the tangent of a double argument (trigonometry):

 - tangent of the angle between lines AC and BC

Let us find k of the line BC from the equality:

18 – 21 k_BC=28+24 k_BC

k_BC= -2/9

Let's make an equation of the side BC:

Equation of the BC side: 2х+9у-65=0

Since BC∩BD at point B, we find the coordinates of point B from the solution of the system:

Knowing the coordinates of point B, write the equation of side AB:

18х+13у-41=0 - equation of side АВ