Question

Determine the value of k so that the line
3x - Ky=8 shall make angle of 45° with the
line x-2y=3.​

Answer 1

$\textbf{Given:}$

$\textsf{Angle between the lines 3x-ky-8=0 and x-2y-3=0 is}\;\mathsf{45^{\circ}}$

$\textbf{To find:}$

$\textsf{The value of k}$

$\textbf{Solution:}$

$\mathsf{Slope\;of\;3x-ky-8=0\;is\;m_1=\dfrac{-3}{-k}=\dfrac{3}{k}}$

$\mathsf{Slope\;of\;x-2y-3=0=0\;is\;m_2=\dfrac{-1}{-2}=\dfrac{1}{2}}$

$\mathsf{Since\;the\,angle\,between\,the\,lines\,is\,45^\circ,\;we\;have}$

$\mathsf{tan\theta=\left|\dfrac{m_1-m_2}{1+m_1\,m_2}\right|}$

$\implies\mathsf{tan45^\circ=\left|\dfrac{\frac{3}{k}-\frac{1}{2}}{1+\frac{3}{k}{\times}\frac{1}{2}}\right|}$

$\implies\mathsf{1=\left|\dfrac{\frac{3}{k}-\frac{1}{2}}{1+\frac{3}{2k}}\right|}$

$\implies\mathsf{\frac{3}{k}-\frac{1}{2}=1+\frac{3}{2k}}$

$\implies\mathsf{\frac{6-k}{2k}=\frac{2k+3}{2k}}$

$\implies\mathsf{6-k=2k+3}$

$\implies\mathsf{6-3=2k+k}$

$\implies\mathsf{3k=3}$

$\implies\boxed{\mathsf{k=1}}$

$\textbf{Find more:}$

Find the value of p, if the straight lines 3x + 7y - 1 = 0 and

7x - py + 3 = 0 are mutually perpendicular.​

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Find p and q, if the following equation represents a pair of perpendicular lines

2x²+ 4xy – py² + 4x +qy+1=0​

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If one of the line given by ax^2+2hxy+by^2=0 is perpendicular to px+ qy=0 then show that ap^2+2hpq+by^2=0

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Answer 2

Answer:

Therefore value of k is 1