Practical lesson 1. Related to the equation of a straight line on a plane
In this lesson, we are going to consider a number of problems related to the equation of a straight line on a plane.
First, we are going to consider an algorithm that allows us to find the angle between two straight lines that are given by common equations. Let's say that there are two straight lines whose equations are shown on the slide, and we need to find the angle between them.
We have a formula that allows us to find the angle between vectors. We also know that the cosine of the angle between straight lines is equal to the modulus of the cosine of the angle either between normal vectors or between guide vectors. I suggest considering normal vectors, given that their coordinates are easily calculated from the general equation. Let us consider the coordinates of the normal vector of the first straight line (18, 6) and the normal vector of the second straight line (5, 10). If we consider a vector that is collinear to the vector with these coordinates, its direction will not change. To make your reading easier, it's best to make the coordinates look simpler. Let us divide the coordinates of the first vector by 6, we get the vector (3, 1) and take it as the normal one. We do the same with the second vector: we divide the coordinates by 5 and get (1, 2). This is the normal vector for the second straight line. Now let's find the cosine of the angle between these vectors.
I denote the angle by φ – this is the angle between the vectors n1 and n2. Using the formula, we multiply them scalarly. 3*1+1*2=3+2 is in the numerator. The denominator is the product of their lengths, that is, the square root of the sum of the coordinate squares (9+1=10) and the length of the second vector is the square root of 1+4 that is 5. Then we calculate: 5 divided by the square root can be written under the common root, and then we take 5 squared out of the root sign. We get 5√2 and reduce it by 5. Then we get rid of the irrationality in the denominator, which gives us the value of √2/2. This is the table value of the cosine, so the angle φ = 45˚. We do not need to use the modulus here, because the cosine is greater than 0, which means that the angle between the vectors is acute. Then it is equal to the angle between our straight lines. Thus, the lines specified in the problem intersect at an angle of 45˚.
Now let's consider a number of problems in which we need to form an equation of a straight line. We have two points A and B, and we need to form the equation of the line AB. The line passes through two given points. We use the canonical equation. Two points A and B form a guiding vector, for example AB, whose coordinates are easy to find: AB(4, -8). Since the coordinates and vectors are not zero, then we write the canonical equation in the following form. As the starting point, we take the point A, then we apply the general formula, and we get (x-2)/4=(y-3)/(-8). We get the canonical equation of the line. It is easy to convert it to a general view. To do this, let's multiply both parts of the equation by 8, then we get 2x-4 on the left (I open the brackets right away), and we have -y + 3 on the right. We transfer everything to the left side and we get 2x+y-7=0. Thus, we get the general equation of the straight line AB.
Now let's make the equation of the midpoint perpendicular to the segment AB. Let me remind you that the midpoint perpendicular is a straight line that passes through the middle of the segment AB and is perpendicular to this segment. The straight line divides the segment in half and forms a right angle with it. Let's use the letter l to mark this line. To write the equation of the line l, note that the guiding vector of the straight line AB is a normal vector for the line l. Then a point, e.g. the midpoint of the segment AB and the normal vector, can set the l line. We can easily find the middle of the segment. Let's do the math. The half-sum of the coordinates, let it be the point M, has coordinates (4, -1), and the vector AB is a normal vector. This means that the equation of a straight line has the following form. The coordinate of the normal vector is multiplied by the difference x-4 plus the second coordinate, i.e. -8, by the difference y-(-1), and equated to 0. We convert the equations to a general form and open the brackets: 4x-8y-16-8. We get the following equation for the mid-perpendicular 4x-8y-24=0.
Another task for making the equation of a straight line. The only difference is that initially a straight line is set by the general equation, and we have the point M that does not lie on the l straight line. We need to make the equation of a straight line that passes through M and is parallel to l.
Let's find the coordinates of the normal vector for the line l – (2, 5). Let's draw a normal vector and set it from the point of the line l as a vector n with coordinates (2, 5). The required straight line is parallel to l, then this vector is normal for it as well, and we again use the formula 2(x+3)+5(y-7)=0. Note that there are the coordinates of the normal vector before the brackets. Inside the brackets, we substitute the coordinates of the point through which the line passes. Then we open the brackets and get the required equation: 2x+5y-29=0.
Now we need a straight line that is perpendicular to the line l. Let's draw. This is the l straight line, there is a point M, and we need a straight line that passes through M and is perpendicular to the l straight line. In this case, the normal vector of the line l for the desired line is the guiding one; and we use the formula for the canonical equation of the line. Since the coordinates of the guiding vector are not equal to zero, we write it as an equation of fractions. The numerator is x+3 and y-7 (on the right). There are the coordinates of the vector 2 and 5 in the denominator. You can write the general equation on your own. We get the canonical equation of a straight line.
Finally, we are going to consider a problem that allows us to find the distance from an arbitrary point on the plane to a given line. Let's revise the concept of the distance. This is the length of the perpendicular from the point M to the line l. Let's mark the base of the perpendicular with the point H. We know the coordinates of the point M, but we do not know the coordinates of the point H, so I marked them with the letters x and y. However, if the point H is on l, then its coordinates satisfy this equation, that is, we have this equation for these numbers. Now we are going to consider the vector HM and denote it with the letter h with an arrow, and h without an arrow stands for the length of this vector. Why do I consider such a vector? The fact is that if we consider the normal vector of the line l, we know that it has the coordinates (a, b), and this vector is collinear to the vector h. It is clear that it does not have to be co-directed with this vector, the direction can be different, but collinearity is required. Thus, the scalar product of the vector n by the vector h is equal to the product of the lengths of these vectors, that is the n vector length is multiplied by the length of the vector h, and the length of the vector h is the required distance. Thus, we get the length of the vector h from this equation; it is the distance MH (the length of the segment MH). We get the scalar product of the vectors divided by the length of the vector n. We can easily find the length of the vector n. This is the square root of the sum of the coordinate squares. It's the same with the scalar product. We need to write here that the vector HM has the coordinates (x0-x, y0-y), and the vector n has the coordinates (a, b). We multiply scalarly, then we multiply the corresponding coordinates and add these products. I multiply and immediately open the brackets: аx0-ax+by0-by, and in the denominator we have the square root of the sum of the coordinate squares of the vector n, that is, √( а2+b2). Now let's group (-ax) and (-by). One more remark. The scalar product can be equal to the product of lengths with a plus sign, or with a minus sign, if we choose a different direction. We do not know which direction the vector n has. Thus, it is more correct to write that the modulus of the scalar product is equal to the product of their lengths, because a minus sign may occur. I add the module sign.
We understand that the distance cannot be less than 0, so there must be only positive numbers everywhere. Now let's go back to our equation. If the minus sign is taken out of brackets, we have аx+by, but ax+by+c=0. h lies on the line l, so the sum of ax+by = -c, and another minus sign, we get a plus sign in the answer. What do we get? We get the following equation. The length of the segment MH is equal to the quotient: the numerator аx0 +by0 remains with the sign of the modulus. The remaining part gives us c. The denominator is the square root of the sum of the squares of the coordinates of the normal vector а2+b2. This formula is useful and we apply it when solving problems. Note that it is easy to remember. The denominator is the length of the normal vector. The numerator is the same expression; just the coordinates of the point M are used instead of variables. It is obvious that if the point M is on a straight line, then all the equation is zero. There is no contradiction. Indeed, the distance from a point lying on a straight line to this straight line is zero. However, if the point M does not lie on the line, this expression is not 0, thus we get a positive number.