If the line ℓ passes through the point P (x0, Y0) and is perpendicular to the line ax + by + C = 0, then the line ℓ equation can be expressed as () A. A（x-x0）+B（y-y0）=0B. A（x-x0）-B（y-y0）=0C. B（x-x0）+A（y-y0）=0D. B（x-x0）-A（y-y0）=0

If the line ℓ passes through the point P (x0, Y0) and is perpendicular to the line ax + by + C = 0, then the line ℓ equation can be expressed as () A. A（x-x0）+B（y-y0）=0B. A（x-x0）-B（y-y0）=0C. B（x-x0）+A（y-y0）=0D. B（x-x0）-A（y-y0）=0

Let the line ℓ perpendicular to the line ax + by + C = 0 be BX ay + M = 0, and the line ℓ passes through the point P (x0, Y0), so - x0B + ay0 = m is substituted into BX ay + M = 0, and the line ℓ: B (x-x0) - A (y-y0) = 0, so D is selected

It is proved that the equation E ^ X-2 = x has at least a point x0 in the interval (0,2), such that e ^ x0-2 = x0

Let f (x) = e ^ x-2-x
Because:
f(0)=1－2－0=－10
If f (x) is continuous on (0,2), then:
There exists x0 ∈ (0,2) such that f (x0) = 0
That is, there exists x0 ∈ (0,2) such that e ^ (x0) - 2 = x0

When x → x0, f (x) is infinite, and limx → x0g (x) = a, it is proved from the definition that when x → x0, f (x) + G (x) is infinite

For any M > 0, ε > 0, there exists δ > 0, when | x-x0 | m, | gx-a | m - | a | - ε, since m and ε are arbitrary, let M1 = m - | a | - ε be any number, that is, for any M1 > 0, | FX + Gx | > M1, so FX + Gx is infinite

Given that the line ax + by + C = 0 (a * 2 + b * 2 is not equal to 0), and passing through the first quadrant and the fourth quadrant, what are the conditions for real numbers a and B?

Let y = 0, then ax + C = 0, x = - C / A
Because the line passes through the first quadrant and the fourth quadrant, - C / a > 0,
That is AC