You will need

- - two lines in the drawing.
- - equation of two straight lines.

Instruction

1

If straight already drawn on the chart, find the solution graphically. This will continue with both or one of the lines so that they intersect. Then mark the intersection point and drop from it perpendicular to the x-axis (usually Oh).

2

With the scale divisions marked on the axis, find the value of x for that point. If it is on the positive axis direction (right from zero), its value will be positive, otherwise – negative.

3

Similarly, find the ordinate of the point of intersection. If the projection point is located above zero – it is positive, below is negative. Write down the coordinates of a point in form (x, y) is the solution of the problem.

4

If direct is specified in the form formula y=KX+b, you can also solve the problem graphically: draw straight on a coordinate grid and find the solution as described above.

5

Try to find a solution to the problem using the given formula. To do this, make these equations system and solve it. If the equation in the form y=KX+b, just Paranaita both parts with x find X. Then substitute the x value into one of the equations and find y.

6

You can find a solution to the method of Kramer. In that case, give the equation to the form А1х+В1у+C1=0 and А2х+В2у+C2=0. According to the Cramer formula x=-(С1В2-С2В1)/(А1В2-А2В1), and y=-(А1С2-А2С1)/(А1В2-А2В1). Please note, if the denominator is zero, then the lines are parallel or coincide, respectively, do not intersect.

7

If you are given straight lines in space in the canonical form, before you start searching for the solutions, check for parallel direct. For this, rate coefficients at t, if they are proportional, such that x=-1+3t, y=7+2t, z=2+t and x=-1+6t, y=-1+4t, z=-5+2t, then the lines are parallel. In addition, direct can interbreed, in this case, the system will have no solutions.

8

If you found out that the lines intersect, find their point of intersection. First Paranaita variables from different direct, conventionally, replacing t on u first and v second-straight. For example, if you direct this x=t-1, y=2t+1, z=t+2 and x=t+1, y=t+1, z=2t+8 you will get an expression of type u-1=v+1, 2u+1=v+1, u+2=2v+8.

9

Express one equation of u, substitute in another and find v (in this problem u=-2,v=-4). Now, to find the point of intersection, substitute the obtained values in place of t (no matter the first or the second equation) and get the coordinates of the point x=-3, y=-3, z=0.