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Given that line L1: ax-y + 2A = 0 and line L2: (2a-1) + ay + a = 0 are perpendicular to each other, find the value of A The solution should be detailed! Ha ha, thank you first
If a = 0
Then y = 0, x = 0
vertical
If a is not equal to 0
ax-y+2a=0
y=ax+2a
(2a-1)x+ay+a=0
y=-(2a-1)/a*x-1
Perpendicular to each other
So a * [- (2a-1) / a] = - 1
2a-1=1
a=1
So a = 0 or a = 1
It is known that the line L1: ax + (1-A) y = 3 and the line L2: (A-1) x + (2a + 3) y = 2 are perpendicular to each other. To find the value of a, I only find a = 3. How can I find a = 1
The necessary and sufficient conditions for the equation to be vertical are as follows: (- A1 / B1) * (- A2 / B2) = - 1 deduces A1A2 + b1b2 = 0 (the left does not consider the case that the denominator is zero, if the denominator is really zero, the formula on the right is also correct, which is the very few true propositions after the condition is relaxed). Since L1 ⊥ L2, a (A-1) + (1-A) (2a + 3)
Given that the lines L1: ax + (1-A) y = 3 L2: (A-1) x + (2a-3) y = 2 are perpendicular to each other, then the value of real number a is?
The lines L1: ax + (1-A) y = 3 and L2: (A-1) x + (2a-3) y = 2 are perpendicular to each other,
∴a(a-1)+(1-a)(2a-3)=0,
∴(a-1)(3-a)=0,
The solution is a = 1 or 3
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