The angle between the line x+y=3 and the line joining the points (1,1) and (-3,4) is

t a n^{- 1} (7)

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t a n^{- 1} (- 1 / 7)

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t a n^{- 1} (1 / 7)

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None of these

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Solution

The correct option is A $t a n^{- 1} (7)$
$y = x - 3; m_{1} = 1 = tan θ_{1}$
$\frac{(y - 1)}{(x - 1)} = \frac{4 - 1}{- 3 - 1}$
$\frac{(y - 1)}{(x - 1)} = \frac{3}{- 4}; m_{2} = \frac{3}{- 4} = tan θ_{2}$
Angle between two lines are $= tan (θ_{2} - θ_{1}) = \frac{tan θ_{2} - tan θ_{1}}{1 + tan θ_{1} + tan θ_{2}} = \frac{m_{2} - m_{1}}{1 + m_{2} m_{1}} = 7 ⟹ | θ_{2} - θ_{1} | = {tan}^{- 1} (7)$

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The angle between the line x+y=3 and the line joining the points (1,1) and (-3,4) is

The correct option is A tan−1(7)y=x−3;m1=1=tanθ1(y−1)(x−1)=4−1−3−1(y−1)(x−1)=3−4;m2=3−4=tanθ2Angle between two lines are=tan(θ2−θ1)=tanθ2−tanθ11+tanθ1+tanθ2=m2−m11+m2m1=7⟹|θ2−θ1|=tan−1(7)