When the line L passes through the point P (2, - 3), the inclination angle is 45 ° larger than that of the line y = 2x-1. The equation of line L is obtained

When the line L passes through the point P (2, - 3), the inclination angle is 45 ° larger than that of the line y = 2x-1. The equation of line L is obtained

Let y = 2x-1 be a, Tana = 2
tan(a+45°) 【tan(A+B) = (tanA+tanB)/(1-tanAtanB),tan45°=1】
=(2+1)/(1-2*1)
=-3
So l: y + 3 = - 3 (X-2)
y=-3x+3

Given that a straight line L passes through a point (1,4), its inclination angle is twice that of the straight line y = 2x + 3, the equation of the straight line L is obtained
How do you write about tan2a (the process of solving tan2a)

The slope of y = 2x + 3 is 2
That is, Tan θ = 2 = = > θ = arctan (2)
The inclination angle of line L is 2arctan (2)
Slope = Tan [2arctan (2)]
= 2tan(arctan2)/[1 - tan²(arctan2)]
= 2(2)/[1 - 2²]
= 4/(- 3)
= - 4/3
So the equation of line L is Y - 4 = (- 4 / 3) (x - 1)
Namely
4x + 3y - 16 = 0
Formula: Tan (2a) = 2tana / (1 - Tan & # 178; a)

It is known that the equation of circle C is x2 + y2-2x + 4y-4 = 0, and the inclination angle of line L is 45 degrees

(x+2)^2+(y-1)^2=5
Center (- 2,1), radius = √ 5
The inclination angle of line L is 45 degrees
Then the slope = tan45 = 1
y=x+b
x-y+b=0
(- 2,1) to straight line = √ 2
So | - 2-1 + B | / √ 2 = √ 2
|b-3|=2
So B = 5, B = 1
So there are two
x-y+5=0
x-y+1=0
What are you asking? It's a linear equation