High School Math Simple Straight Line Questions, High School Math Straight Line Problems?

When AB is on both sides of the straight line L, L passes through the midpoint of AB M coordinates (2, 3) MA=MB=2, and the distance from A to the straight line is 1, so the angle between L and the straight line AB is 30° and the slope of the straight line AB is k= 3, so the tilt angle of L is 30° or perpendicular to the X axis (it is more clear to see the drawing), and L passes through the point M

The equation for l is y- 3=(x-2) 3, i.e. x- 3y-1=0 or x=2

When ab is on the same side of the line l, l is parallel to the straight line ab, let the equation be y= 3x+t and the distance from the point to the straight line: 1=|√3-0+t|2, solve t=- 3 2 at this time, the linear l equation is y= 3x+2- 3, or y= 3x-2- 3, a total of four.

Straight line: The one that extends infinitely at both ends is called a straight line.

First of all, if the slope does not exist, it is clear that there is x=2

Then see if the perpendicular bisector between the two points meets the requirements, |ab|=4 does not equal 2 and therefore does not match.

In this way, the straight lines that meet the requirements of the problem are on both sides of the AB line, and the slope parallel to AB = (2 3-0) (3-1) = 3

Let the straight line be y= 3x+b then |√3+b|(3+1)=1 b=2-3 or b=-2-3

The equation for a straight line is 3x-y+2- 3=0 or 3x-y-2- 3=0 or x=2

You connect AB and consider the case where A and B are on the same side as L or in an allopathy.

It could be two straight lines parallel to AB, or two lines at an angle of 30 degrees to it

1) The slope does not exist, apparently there is x=2

2) or. The distance of l is equal to 1, ab parallel straight line l

ab linear two-point formula a(1,0),b(3,2 3)ab: 3x-y- 3=0

Tilt angle a = 60°

The distance is equal to 1Intercept in the y-axis = - 3 2

So: straight line l:

3x-y-√3±2=0，x=2

For this question, you should choose different methods based on your understanding of the knowledge points.

Fang 1 (simple thinking, easy calculation).

Once you know the intersection of two straight lines, you can first find the intersection point, then set the straight line equation at a certain point, and then use the distance formula d=3 to get the unknown, and then get the straight line equation.

Fang 2 (requires a deep understanding).

Let's take an example, if there is a straight line ay=bx+c, the line passes a certain point (m, n), and moves the function to one side to get ay-bx-c=0, then when x=m, y=n satisfies the equation equal to zero.

Brought into this problem, the straight line l must pass the fixed point, then, it must be ensured that the sub must be equal to zero after bringing this point into the equation, and l1 and l2 also meet the point after bringing in zero, so the straight line l must include the conditions that l1 and l2 satisfy, so the straight line l can be set to be 3x-2y+4+k(2x+y-16)=0, and then the condition that the distance is equal to three can be solved.

k<=-3 or k>=0, the problem can be transformed into a straight line l and the line segment lead ab has an intersection of the mountain pin, find the slope of l, then the straight line l can be a straight line between the straight line pa and the straight line pb through the p point, k is less than the slope of pa or greater than the slope of pb.

It's easy to draw.

After p(0,1) Gaona and a(1,-2) to make a closed mind divination line, find the straight line a through p(0,1) and b(2,1) to make a straight line, find the straight line b, and find the inclination angle and slope of the two straight lines of a and b obtained by the straight line b.

It's a very simple one-god math problem....Let's start with a picture....Make three points as required, connect and then make a straight line, make a straight line in the pass, and find the slope k of these two straight lines respectively, and these two slopes are the range sought....Because it can be seen from the figure that the straight line passing through the p point and the line segment ab have a constant intersection point, then the intersection point can only reach b at the top, and the bottom can only reach aSo that's the range....The answer is to ask for your own swimming....It's simple....

y'=3x^3

3x 3=4 (the slope of x+4y+1=0 is -1 4, and the slope of the straight line perpendicular to it is 4).

x cube root (4 3).

So the tangent is ((4 3) 1 3, (4 3) 4 3) l and the equation is (y-(4 3) 4 3) = 4(x-(4 3) 1 3).

From known, the slope of the tangent is 4

Again'=4x^3

Set the tangent coordinates to (x,x 4).

Then 4x 3=4 gives x=1, then the tangent coordinates are (1,1) so the equation for l is: y-1=4(x-1).

Let the inclination angle of the straight line x-4y+3=0 be a

Then the inclination angle of the straight line is 2a

tga=1/4

tg2a=2tga/(1-tga^2)=2*(1/4)/(1-1/16)=8/15

So the slope of the new line is 8 15

According to the point slope: y-2=(x-3)*8 15

8x-15y+6=0

4y=x+3 y=1 4 y+3 4 oblique angle tan = 1 4 =arctan 1 4

2arctan 1 4 The slope of the straight line k=tan 2arctan 1 4

A straight line passes through (3,2) y-2=tan 2arctan1 4 (x-3).

4y=x+3 y=1 4 y+3 4 oblique angle tan = 1 4 The straight line angle through the point p is 2a, then tan2a=8 15

The slope of the straight line k=tan 2a=8 15 passes through (3,2) y-2=8 15 (x-3).

This question should be an examination of the application of the chamfer formula.

If we know that the slope of the line L1 is K1 and we know that the slope of the line L2 is K2, find the slope K3 of the symmetrical line L3 of the line L1 with respect to the line L2.

Get: (k2-k3) (1+k2·k3)=(k1-k2) (1+k1·k2).

The straight line to be found in the problem can be regarded as a symmetrical straight line on the x-axis about the known straight line, and the x-axis can be regarded as a straight line with a slope of 0, and the above formula code can be used to obtain (1 4-k3) (1+k3) =(0-1 4) (1+0), k3=8 15, and the equation of the straight line to be found is y-2=8 15(x-3), which is reduced to y=8 15x+2 5

If the slope does not exist, it means that the slope is infinite, and the line is perpendicular to the x-axis, so the coefficient of x in the equation is not equal to 0, and the coefficient of y is equal to 0, which is solved by 4-m 2=0 to m=2 or -2

When m=2, the coefficient of x 2+m-m 2 is 0, which does not match the meaning of the title, so m can only take -2 without consideration

The slope should not exist should be a zero denominator, and the one to the right of the equal sign in your second line should not have a negative sign.

Let -(4-m 2)=0 solve m = plus or minus 2, and substitute plus or minus 2 into the original formula, m=2 does not match, so m = -2

Solution: Let the equation for this line be y-0=k(x-3).

i.e. y=kx-3k

The intersection point with l1 is: (x1,y1).

2x1-y1-2=0

y1=kx1-3k

The intersection of x1=(2-3k) (2-k) y1=-4k (2-k) and l2 is: (x2,y2).

x2+y2+3=0

y2=kx2-3k

x2=(3k-3) (k+1) y2=-6k (k+1)Since point p is the midpoint of the line segment, so: 3=[(2-3k) (2-k)+(3k-3) (k+1)] 2

0=(-6k (k+1)+(4k) (2-k)) 2k=8 The linear equation is: y=8x-24

The equation for the straight line system passing through the p point is: y=k(x-3).

The abscissa of the intersection points obtained by combining with equations 2x-y-2=0 and x+y+3=0 are respectively:

3k-2) (k-2) and (3k-3) (k+1) since p is the midpoint of these two intersections.

3k-2) (k-2) + (3k-3) (k+1) = 3 2 passes are sorted out, and k = 8 is obtained

So the equation is: y=8(x-3).

A false: the slope does not necessarily exist, it can not exist;

B false: After the tilt angle is large, the slope may be negative, but it will be smaller.

C is true and d is false, and the slope may not exist, such as when the tilt angle = 90 degrees.

A is not true, there is no slope of a straight line perpendicular to the x-axis.

b is not true, because the tangent of the tilt angle is the slope and the tangent function is a periodic function, so the tangent of an obtuse angle over 90° is not as large as the tangent of an acute angle.

c No, it can be 180°

d right. The tangent of the same angle is the same, so the slope is also equal.

c The tilt angle is +2k or 0

d is wrong, when the angle is 90 degrees, there is no slope, there is no comparison.

When the slope is 0, x=1, the distance from the point p(2,2) to the straight line x=1 is 3. This can be drawn directly, and the relationship between the dots and the line is very clear. Because x=1 is a perpendicular axis.