Problems with linear equations, problems with linear equations?

The so-called straight line system refers to all straight lines that pass through a fixed point.

The linear equation is to use the method of undetermined coefficients to find a specific straight line at a certain point.

For example, find the intersection of two straight lines with a slope of 5 L3L1:0=X-Y

l2：0=2x-3y+1

Let l3 be: (2x-3y+1)+ x-y)=02x+ x-3y- y+1=0

2+λ)x-(3+λ)y+1=0

3+λ)y=(2+λ)x+1

y=(2+)x(3+)1(3+) Therefore, the slope is:

Back generation l3,2-13 4)x-(3-13 4)y+1=05x+y+4=0

Then, l3 is: 5x-y-4=0

Let's check it out:

The intersection of l1 and l2 is (1,1) (the simultaneous equation is obtained) and substituting (1,1) into l3,5-1-4=0 is true, then it means that l3 passes the intersection of l1 and l2.

The slope of the apparent L3 is 5

are eligible.

If you don't understand, please ask.

It may be used to find straight line equations by the method of undetermined coefficients, and it may be applied to the Lagrangian multiplier method to find the conditional extrema of binary functions.

A straight line passes a fixed point.

That is, a straight line passes through the intersection of these two straight lines.

Since the known point is (0,-1),-1)=1, we get the part of 1 of k+1.

Please review the formula for the distance between a point and a straight line.

Isn't that what the point-to-straight distance formula is all about?

There is a mistake in the red pen draft of this question!

The lower left corner should be x=4x +4x+1, so option A is substituted δ not 0

In this question, d should be substituted and found to be 0, so d is correct.

Untie. When the intersection of the line l and the x-axis is on the positive semi-axis, i.e., x 0, and the intersection with the y-axis is on the positive semi-axis of the y-axis, i.e., y 0, the straight line is no more than the third quadrant.

m+2)x+(m-1)y-2m+3=0

When y 0, x (2m-3) (m+2) 0 when x 0, x (2m-3) (m-1) 0 if 2m-3 0, i.e., m3 2, m+2 2+3 2 0, m-1 3 2-1 0, if 2m-3 0, i.e., m3 2, m+2 0, m-1 0, i.e., m-2

i.e. m 3 2 and m -2, the straight line is no more than the third quadrant.

The slope of the straight line l k=-(m+2) (m-1).

If the straight line l is not the third quadrant.

The slope k of the (0,0) point and (0,0) point is reversed around the fixed point from the parallel x-axis to the (0,0) point'=-7

Then the slope of l is from 0 to -7

0≤k≤-7

0≥-(m+2)/(m-1)≥-7

0≤(m+2)/(m-1)≤7

There's something to be done.

Another way: know the coordinates of the three points of abc and use two of them to find an equation, such as using a and b two points to find an equation.

The equation of the straight line where ab is located and the length ab is found by using the coordinates of two points, and then the distance d from point c to the straight line is found by using the formula of the distance from point c to the straight line ab, that is, the height on the edge of ab.

then the area is ab *d 2

The method hopes! If you don't understand, please ask me again.

x=2 represents a straight line that is parallel to the y-axis and perpendicular to the x-axis at the crossing point (2,0). The general equation for a straight line is y=kx+b, but there are two special cases, a straight line parallel to the y-axis, which is represented by x=; A straight line parallel to the x-axis, denoted by y=. The equal sign is followed by a number, e.g. y=1 is a straight line perpendicular to the x-axis through the point (0,1).

First you turn the circle into a standard equation of x 2+(y+1) 2=4, then the center of the circle is (0,-1) and the radius is 2. Ab is required, and the Pythagorean theorem can be used to distance from a point to a straight line.

The distance from the origin to the straight line ab is 2 2 1 2 = 2, then 1 2ab=(4-2) 1 2= 2, then ab=2 2.