If the slope of line L1 is 2, L2 passes through point P (0,2) Q (x, 10), L1 is parallel to L2, then log2x =?

If the slope of line L1 is 2, L2 passes through point P (0,2) Q (x, 10), L1 is parallel to L2, then log2x =?

L1 is parallel to L2
The slope of L2 is 2
y-2=2x
10-2=2x
x=4
log2x=log24=2

It is known that the inclination angle of the line L1: y = MX is twice that of the line L2: y = NX, and the slope of L1 is four times that of L2. The product Mn of non-zero real numbers m, n is obtained

The tilt angle cannot be greater than 90 degrees
Because the slope is positive and negative, if one inclination angle is greater than 90, the other must be greater than 180, which does not meet the definition of inclination angle
Let L1 tilt angle a, l22a, obviously the angle between two straight lines is a
The slope angle formula, Tana = | M-N | / (1 + Mn), M = 4N, Tana = n
tana=3tana/(1+mn)
mn=2

The slopes of lines L1 and L2 are - 1A, − 23 respectively. If L1 ⊥ L2, then the value of real number a is ()
A. -23B. -32C. 23D. 32

∵ L1 ⊥ L2, ■ − 1a × (− 23) = - 1, the solution is a = − 23

It is known that the slopes of line L1 and line L2 are m and N respectively. If the inclination angle of line L1 is twice that of line L2, the slope of line L1 is four times that of line L2, and line L2 is not parallel to the x-axis, the values of M and N can be obtained

If the inclination angle of L2 is θ (≠ 0), then the inclination angle of L1 is 2 θ, so m = Tan (2 θ) = 2tan θ / (1 - (Tan θ) ^ 2) = 2n / (1-N ^ 2) = 4N, the solution is n = √ 2 / 2 or n = - √ 2 / 2 (rounding off). Why rounding off the negative value? This is because when n = - √ 2 / 2, θ is an obtuse angle, then 2 θ exceeds 180 degrees