If line x - y -1-0 is perpendicular to the line 2 x - y -1-0 and parallel to the line x -3 y -1-0- then the value of - is

Let the lines be nbsp; l_1: x+α y+1=0 l_2: nbsp;2x β y+1=0 l_3: nbsp;x (β 3)y 1=0 Slope of the lines are nbsp; m_1 = 1α, m_2 =2β, m_3 = 1β 3 From ...

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Standard XII

Mathematics

Angle between Two Lines

If line x +α ...

Question

If line $x + α y + 1 = 0$ is perpendicular to the line $2 x - β y + 1 = 0$ and parallel to the line $x - (β - 3) y - 1 = 0$ , then the value of $| α - β |$ is

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Solution

Let the lines be
$l_{1} : x + α y + 1 = 0 l_{2} : 2 x - β y + 1 = 0 l_{3} : x - (β - 3) y - 1 = 0$
Slope of the lines are
$m_{1} = - \frac{1}{α}, m_{2} = \frac{2}{β}, m_{3} = \frac{1}{β - 3}$
From the given conditions,
$m_{1} \times m_{2} = - 1 \Rightarrow - \frac{2}{α β} = - 1 \Rightarrow α β = 2 \dots (1) m_{1} = m_{3} \Rightarrow - \frac{1}{α} = \frac{1}{β - 3} \Rightarrow α + β = 3 \dots (2)$
Now,
$| α - β | = \sqrt{(α + β)^{2} - 4 α β} \Rightarrow | α - β | = \sqrt{9 - 8} = 1$

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Angle between Two Lines

Standard XII Mathematics

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If line x+αy+1=0 is perpendicular to the line 2x−βy+1=0 and parallel to the line x−(β−3)y−1=0, then the value of |α−β| is

Let the lines be l1:x+αy+1=0l2:2x−βy+1=0l3:x−(β−3)y−1=0 Slope of the lines are m1=−1α, m2=2β, m3=1β−3 From the given conditions, m1×m2=−1⇒−2αβ=−1⇒αβ=2⋯(1)m1=m3⇒−1α=1β−3⇒α+β=3⋯(2) Now, |α−β|=√(α+β)2−4αβ⇒|α−β|=√9−8=1

If line $x + α y + 1 = 0$ is perpendicular to the line $2 x - β y + 1 = 0$ and parallel to the line $x - (β - 3) y - 1 = 0$ , then the value of $| α - β |$ is