If the inclination of the straight line L is α=∏/4, its slope is equal to?

Slope =tanα=tan(π/40)= 1

So the slope is

1

2.

When the straight line L passes through the point A (- 1, 2) B (2, 3), the slope k of the straight line L is equal to?

The slope is

k =(2-3)/(- 1-2)=(- 1)/(-3)= 1/3

3.

What are the coordinates of the intersection of the line 3x-y- 1=0 and the line x-2y+3=0?

3x-y- 1=0

& lt 1 & gt；

x-2y+3=0

& lt2 & gt

Then <1> * 2-<; 2>, eliminate y to get: 5x-5=0.

x= 1

Substitute x= 1.

3x-y- 1=0

, get: 3-y- 1=0.

That is, 2-y=0 and y=2.

So the intersection coordinates are

( 1,2)

4.

The distance from the point m (-2,3) to the straight line 3x-4y+8=0 is equal to?

Use the formula:

d=|AXo+BYo+C|

/

√(A？ +B？ )

Get:

The distance is

d=|-2*3+(-4)*3+8|

/

√(3? +4? )

=|-6- 12+8|/√25

= 10/5

=2

So the distance from m (-2,3) to the straight line 3x-4y+8=0 is equal to

2

5.

Given that the straight line l 1 passes through the point (4,-1) and is parallel to the X axis, what is the linear equation of l 1?

Parallel to the x axis, the equation is

y=- 1

According to the following conditions, the linear equations are solved respectively and converted into general linear formulas.

The inclination of (1) intersection (1, -4) is 30 degrees.

Slope k = tan 30 = √ 3/3

Then y=√3/3(x- 1)-4.

Into a general formula:

x-√3y- 1-4√3=0

(2) Pass through point A (1, 2)

b(0.5)

get through

B (0 0,5), the equation can be set as

y=kx+5

A( 1, 2) is on a straight line, then

2=k+5

Then k=-3

therefore

y=-3x+5

Turn it into a general formula

3x+y-5=0

(3) Crossing point (-2,3) and parallel to the straight line 3x-4y+ 1=0.

3x-4y+ 1=0 This line is

y=3/4x+ 1/4，

The slope is

3/4,

Then the slope of the straight line passing through point (-2,3) is also.

3/4

So the equation is

y=3/4(x+2)+3

That is y=3/4x+ 15/4.

Into a general formula:

3x-4y+ 15=0