When a changes in the interval (0,2), find the minimum value of the quadrilateral enclosed by the line and 2 coordinates

When a changes in the interval (0,2), find the minimum value of the quadrilateral enclosed by the line and 2 coordinates

[analysis] because the line L1 and L2 both pass through the fixed point (2,2), and the intercept of line L1 on the y-axis is B1 = 2-A > 0, and the intercept of line L2 on the x-axis is B2 = A2 + 1 > 0, so s = B1.2 + · b2.2 = a2-a + 3 = (a-0.5) ^ 2 + 2.75. When a = 0.5, s is the smallest. [answer] 0.5
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1. If the line ax + 2Y + 2 = 0 is parallel to the line 3x-y-2 = 0, then the coefficient a=
2、
2. The image of function y = SiNx is translated according to vector a = (- π / 2,2) and coincides with the image of function g (x), and the expression of G (x) is obtained
3. Given that the line L is perpendicular to the line 3x + 4y-5 = 0, and the chord length cut by the circle xsquare + ysquare + 4Y = 0 is equal to 2 * root sign 3, then the equation of the line L is

If the line a1x + b1y + C1 = 0 is parallel to the line a2x + b2y + C2 = 0, there must be a relationship: a1b2-a2b1 = 0 because the line ax + 2Y + 2 = 0 is parallel to the line 3x-y-2 = 0, so a × (- 1) - 3 × 2 = 0, the solution is a = - 62

The necessary and sufficient condition of "the line: x + (A-1) y + 1 = 0 is parallel to the line: ax + 2Y + 2 = 0" is______ ．

When a = 1, the slope of the line x + (A-1) y + 1 = 0 does not exist, the slope of the line ax + 2Y + 2 = 0 is - 12, and the two lines are not parallel. When a ≠ 1, the line x + (A-1) y + 1 = 0 is transformed into y = 11 − ax + 11 − a, and the line ax + 2Y + 2 = 0 is transformed into y = − a2x − 1. If the line x + (A-1) y + 1 = 0 is parallel to the line ax + 2Y + 2 = 0, the slopes of the two lines are equal and the intercept is not equal There is 11 − a = − a211 − a ≠ & nbsp; − 1, the solution is a = - 1, so the necessary and sufficient condition of "the line: x + (A-1) y + 1 = 0 is parallel to the line: ax + 2Y + 2 = 0" is a = - 1. So the answer is a = - 1

If the line ax + 2Y + 2 = 0 is parallel to the line 3x-y-2 = 0, then the value of a is?

a/3 = 2/(-1)
Yide a = - 6
When solving parallel problems, if there are multiple solutions, do not forget to substitute the numerical value to see if there is coincidence

If the line ax + 2Y + 2 = 0 is perpendicular to the line 3x-y-2 = 0, then the value of a is ()
A. −32B. 32C. −23D. 23

If ax + 2Y + 2 = 0 is perpendicular to 3x-y-2 = 0, then the slopes of the two lines exist, and the product of the slopes is - 1, ■ − A2 × 31 = − 1. The solution is a = 23, so D

If the line ax ‐ 2Y + 2 = 0 is parallel to the line 3x-y-2 = 0, then the coefficient a is equal to ()
A. 6B. -3C. -32D. 23

∵ ax ‐ 2Y + 2 = 0 is parallel to the line 3x-y-2 = 0, ∵ 3A = − 1 − 2 ≠− 22, the solution is a = 6, so the answer is: 6

If the line ax ‐ 2Y + 2 = 0 is parallel to the line 3x-y-2 = 0, then the coefficient a is equal to ()
A. 6B. -3C. -32D. 23

∵ ax ‐ 2Y + 2 = 0 is parallel to the line 3x-y-2 = 0, ∵ 3A = − 1 − 2 ≠− 22, the solution is a = 6, so the answer is: 6

If the line ax-2y + 1 = 0 is perpendicular to the line 3x + Y-2 = 0, then a is

y=ax/2+1/2 y=-3x+2
The slopes are reciprocal
a/2*(-3)=-1
a=2/3

A = 0 is a straight line ax-2y-1 = 0 and 3x + 2 = 0 perpendicular to what conditions

a=0
Is - 2y-1 = 0
3x+2=0
Vertical, so it's a sufficient condition
And if it's vertical
3x + 2 = 0 vertical X-axis
So ax-2y-1 = 0 is perpendicular to the Y axis
So a = 0
So it's a necessary condition
So it's a necessary and sufficient condition

If the line x + 2y-1 = 0 is perpendicular to ax-y + 1 = 0, then the value of real number a is

Y = (- x + 1) / 2 gives k = - 1 / 2
Another y = ax + 1 gives k = a
Because they are perpendicular to each other, so - 1 / 2 x a = - 1
That is, a = 2