Advice 1: How to write the equations of the sides of the triangle

There are many ways to define a triangle. In analytic geometry one of these methods is to specify the coordinates of its three vertices. These three points determine a triangle uniquely, but for completeness we still have to make the equations of the sidesconnecting the vertices. Instruction
1
You are given the coordinates of three points. Denote them as (x1, y1), (x2, y2), (x3, y3). It is assumed that these points are the vertices of some triangle. The task is to write the equations of its sides, or rather the equations of those straight, where are these parties. These equations should be:
y = k1*x + b1;
y = k2*x + b2;
y = k3*x + b3.Thus, you have to find the angular coefficients k1, k2, k3 and offset b1, b2, b3.
2
Make sure that all points are distinct. If any two match, then the triangle degenerates into a line segment.
3
Find the equation of the line passing through the points (x1, y1), (x2, y2). If x1 = x2, then the required straight vertical and its equation is x = x1. If y1 = y2, the direct horizontal and its equation is y = y1. In the General case, these coordinates will not be equal to each other.
4
Substituting the coordinates (x1, y1), (x2, y2) in the General equation of a straight line, you will get a system of two linear equations:k1*x1 + b1 = y1;
k1*x2 + b1 = y2.Subtract one equation from the other and solve the resulting equation for k1:k1*(x2 - x1) = y2 - y1 therefore, k1 = (y2 - y1)/(x2 - x1).
5
Substituting the expression in any of the original equations, find an expression for b1:((y2 - y1)/(x2 - x1))*x1 + b1 = y1;
b1 = y1 - ((y2 - y1)/(x2 - x1))*x1.Since we already know that x2 ≠ x1, we can simplify the expression by multiplying y1 (x2 - x1)/(x2 - x1). Then for b1 you will get the following expression:b1 = (x1*y2 - x2*y1)/(x2 - x1).
6
Check is the third of the given points to the found line. To do this, substitute the values of (x3, y3) in the derived equation and see whether they are equal. If it is observed, therefore all three points are collinear, and the triangle degenerates into a line segment.
7
In the same way as described above, output equation for a straight line passing through the points (x2, y2), (x3, y3) and (x1, y1), (x3, y3).
8
The final form of the equations for the sides of a triangle given coordinates of vertices, as follows:(1) y = ((y2 - y1)*x + (x1*y2 - x2*y1))/(x2 - x1);
(2) y = ((y3 - y2)*x + (x2*y3 - x3*y2))/(x3 - x2);
(3) y = ((y3 - y1)*x + (x1*y3 - x3*y1))/(x3 - x1).

Advice 2 : How to find the height of a triangle if the coordinates of the points

The height of the triangle is called a straight line connecting the top of the figure with the opposite side. This cut must be perpendicular to the side, so each vertex can hold only one height. Because the peaks in this figure three, heights it the same. If a triangle specified by coordinates of its vertices, the calculation of the length of each of the heights can be produced, for example, using the formula for finding the area and calculating the lengths of the sides. Instruction
1
Assume in the calculations that the area of a triangle is equal to half of a work the length of either of the parties on the length of the height lowered on this side. From this definition it follows that in order to find the height you need to know the area of the shape and the length of the side.
2
Start with calculation of the lengths of the sides of the triangle. Label the coordinates of the vertices: A(X₁,Y₁,Z₁), B(X₂,Y₂,Z₂) and C(X₃,Y₃,Z₃). Then the length of the side AB, you can calculate by the formula AB = √((X₁-X₂)2 + (Y₁-Y₂)2 + (Z₁-Z₂)2). For the other two sides of these formulas are as follows: BC = √((X₂-X₃)2 + (Y₂-Y₃)2 + (Z₂-Z₃)2) and AC = √((X₁-X₃)2 + (Y₁-Y₃)2 + (Z₁-Z₃)2). For example, for a triangle with coordinates A(3,5,7), B(16,14,19) and C(1,2,13) the length of the side AB will be √((3-16)2 + (5-14)2 + (7-19)2) = √(-132 + (-92) + (-122)) = √(169 + 81 + 144) = √394 ≈ 19,85. The lengths of the sides BC and AC, calculated in the same way, will be equal √(152 + 122 + 62) = √405 ≈ 20,12 and √(22 + 32 + (-62)) = √49 = 7.
3
Knowledge of the lengths of the three sides, obtained in the previous step, it is enough to compute the area of the triangle (S) by Heron's formula: S = ¼ * √((AB+BC+CA) * (BC+CA-AB) * (AB+CA-BC) * (AB+BC-CA)). For example, after substituting in this formula the values obtained from the coordinates of the trianglea sample from the previous step, this formula will give a value of S = ¼ *√((19,85+20,12+7) * (20,12+7-19,85) * (19,85+7-20,12) * (19,85+20,12-7)) = ¼ *√(46,97 * 7,27 * 6,73 * 32,97) ≈ ¼ *√75768,55 ≈ ¼*275,26 = 68,815.
4
Based on the area of the trianglecalculated in the previous step, and the lengths of the sides obtained in the second step, calculate the height for each of the parties. Since the area is equal to half the height works for the length of the side to which it is held, to find the height divide the double area for length right side: H = 2*S/a. For of the example used above the height lowered on the side AB will be 2*68,815/16,09 ≈ 8 and 55, the height to the side of the sun will have a length of 2*68,815/20,12 if 6,84, but for AC, this value will be equal to 2*68,815/7 if of 19.66.

Advice 3 : As the coordinates of the triangle vertices to find equation of sides

In analytic geometry a triangle in the plane can be specified in a Cartesian coordinate system. Knowing the coordinates of the vertices, you can write the equations of the sides of the triangle. These will be the equations of three lines that intersect, form a shape. You will need
• - handle;
• paper for records;
• calculator.
Instruction
1
Video on the plane described by the equation: ax+by+C = 0, where x,y – coordinates along axis 0x, and the axis 0у any point of the line; a, b, C – numerical coefficients. Moreover a and b cannot be zero simultaneously. This type of entry is called the General equation of a straight line.
2
Direct you can specify an expression of the form: y = kx+c. This is the equation of a line with the angular coefficient k, which is the tangent of the angle formed by the intersection of this line with axis 0x.
3
Knowing the coordinates of two points A (x1, y1), (x2;Y2), you can write the equation of a straight line drawn through these points, using the ratio: (u-U1)/(U1-U2)=(x-x1)/(U1-U2). Then, transforming this equation, bring it to the form as in step 1 or 2.
4
Consider the algorithm of solving the problem on a concrete example. Given three vertices of a triangle with known coordinates: A (9;8), (7;-6), (-7;4). Write the equation of direct forming it. 5
Find the equation for the line AB. Apply the formula from step 3, substituting values of the coordinates of points A and b: (-8)/(8-(-6)) = (x-9)/(9-7). Convert it: (y-8)/14 = (x-9)/2 or 2(y-8) = 14(x-9). Reduce the equation by dividing the left and right sides by two, expand the brackets: y = 7x-63+8 = 7x-55.
The equation for AB is: y = 7x-55. Or: 7x-55 = 0 (AB).
6
Similarly, write the equation for direct sun: (-(-6))/(-6-4) = (x-7)/7-(-7)). (at+6)/(-10) = (x-7)/14. 7(y+6) = -5(x-7). 7U+42 = -5x+35. 7U = -5x-7. y = -5/7x-1.
The equation for m: y = -5/7x-1. Or: -5x-7U-7 = 0 (Sunday).
7
Then the equation for a straight SA: (u-8)/(8-4) = (x-9)/(9-(-7)). 16(y-8) = 4(x-9). 4U-32 = x-9. 4U = x-9+32. y = 0.25 x+5,75.
The equation for CA: y = 0.25 x+5,75. Or: x-4U+23 = 0 (SA).
8
You made the equation three sides of the figure. For the self build triangle in the coordinate system. Locate on the drawing the values of the intersections of straight lines with the axis 0у. Compare these coordinates with those obtained in the equation. For example, for a (BC) when y = 0, x = -1,4.
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