1 Correct your mistake. The slope of the equation is not y 1-y2=k(x 1-x2), but y-y 1=k(x-x 1), that is to say, the slope is k, and it passes through the fixed point. The equation of y=kx+b is called truncated type, which represents a straight line with a slope of k and an intercept of b. As for y 1-y2=k(x 1-x2), when the straight line is not perpendicular to the X axis, it is calculated from two points on the known straight line (x 1, y/kloc-0).
Excuse me, how do you use these two kinds? Which topic uses the first one and which topic uses the second one?
B When calculating the slope, should we look at k in y=kx+b or k=y 1-y2/x 1-x2?
These two issues are closely related. Let me answer it together.
A If the equation of a straight line is given directly, and it is in the form of y=kx+b, then its slope can be directly known as K. ..
When you know the coordinates of two points on a straight line that is not perpendicular to the X axis, you need to use the formula k=y 1-y2/x 1-x2 to find the slope or equation of this straight line.
B with the above answer, if the equation of a straight line is known to be in the form of y=kx+b, then its slope is K.
When you know the coordinates of two points on a straight line that is not perpendicular to the X axis, you need to use the formula k=y 1-y2/x 1-x2.
3 For example, if the equation of the straight line L is known as y=x+ 1, then the inclination of the straight line L is?
This problem is ①k=y/x+ 1= 1, so it is 45. ② Look directly at k= 1 before the unknown x in the equation, so it is 45.
Is the solution of ① right or ② right?
The answer is right, but the solution is wrong. The title gives the form of y=kx+b, where k=b= 1, and its slope should be k= 1. If you follow your solution, then y=kx+b gives y/kx+b= 1, then the slopes of all straight lines are 1.
Where k≠y/x+ 1 should be k = y-1/x.
There is an inaccurate concept in ②. X should be called an independent variable rather than an unknown. This solution is correct.
4 For example, if points (2,0) intersect, the linear equation with a slope of 3 is (d).
A.Y = 3x+2b。 Y = 3x-2c。 Y = 3 (x+2) D. Y = 3 (x-2) Why does this question not look at k=3 in front of X but move right and left to get k=3?
Sorry, I really don't understand what you think. What does "k=3 before x" mean? What's on the right? What's on the left? Move what from the right to the left?
Now that we know the slope and a point on a straight line, the correct thinking of this problem is to use what you said, y 1-y2=k(x 1-x2), and I will correct it for you, that is, the point is oblique, that is, y-y1= k (x-x/kloc-).
K=3 x 1=2 y 1=0。 If you substitute the formula of point inclination, it is y=3(x-2).
Oh, I'm exhausted.