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KEAM 2010 Math Paper

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Question : 115 of 120

Marks: +1, -0

The slopes of the lines which make an angle

45^{\circ}

3x - y = -5

are

1,-1
$\;\frac{1}{2}, -1$
$1, \;\frac{1}{2}$
$2, -\;\frac{1}{2}$
$-2, \;\frac{1}{2}$

Solution:

The slope of line

3x - y = -5

is

m_1 = 3

∴ The slope of required line is

= \;\frac{m_{1} \pm \tan 45^{\circ}}{1 \mp m_{1} \cdot \tan 45^{\circ}} = \;\frac{3 \pm 1}{1 \mp 3 \cdot 1}

= -2, \frac{1}{2}

Alternate

Given line

3x - y = -5

m_{1} = 3

Let slope of another line

= m_{2}

Now, angle between both the lines

\tan \theta = \left| \;\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} \right|

\tan 45^{\circ} = \;\frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}

or

\;\frac{m_{2} - m_{1}}{1 + m_{1} m_{2}}

(If

m_{1} > m_{2})

\Rightarrow \;\; 1 + m_{1} m_{2} = m_{1} - m_{2}

\Rightarrow \;\; 1 + 3 m_{2} = 3 - m_{2}

\Rightarrow \;\; m_{2} = \;\frac{1}{2}

(If

m_{2} > m_{1})

Then,

1 + 3 m_{2} = m_{2} - 3

\Rightarrow \;\;\; m_{2} = -2

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