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  • The tangent of angle between the lines whose intercepts on the axes are a, minus b and b, minus a, respectively, is A. a square minus b square divided ab B. b square minus a square divided 2 C. b square minus a square divided 2ab D None of these

The tangent of angle between the lines whose intercepts on the axes are a, – b and b, – a, respectively, is
A. \frac{a^{2}-b^{2}}{ab}
B. \frac{b^{2}-a^{2}}{2}
C. \frac{b^{2}-a^{2}}{2ab}
D. None of these

Answers (1)

First equation of line in intercept form=

\frac{x}{a}+\frac{y}{-b}=1

\frac{x}{a}-\frac{y}{b}=1

  bx-ay=ab……. (i)

   Let the second equation of line having intercepts on the axes b, -a is  \frac{x}{b}+\frac{y}{-a}=1

\frac{x}{b}-\frac{y}{a}=1

  ax-by=ab…..(ii)  

 Now we find the slope in the first equation  bx-ay=ab 

  ay=bx-ab  

y=\frac{b}{a}x-b

  Slope of the equation m_{1}=\frac{b}{a}

 Now, we find the slope of equation (ii) ax-by=ab 

 by=ax-ab 

y=\frac{a}{b}x-a

 Slope of the equation m_{2}=\frac{a}{b}

  Let θ be the angle between the given two lines \tan \theta =\left |\frac{\left ( m_{1}-m_{2} \right )}{1+m_{1}m_{2} } \right |

 Putting the values of m1 and m2 in above equation, we get \tan \theta =\left |\frac{\left ( \frac{b}{a}-\frac{a}{b} \right )}{1+\left ( \frac{b}{a} \right )\left (\frac{a}{b} \right )} \right |

  \tan \theta =\left | \frac{\frac{b^{2}-a^{2}}{ab}}{1+1} \right |=\left | \frac{\left (b^{2}-a^{2} \right )}{2ab} \right |=\left | \frac{b^{2}-a^{2} }{2ab} \right |  is the required angle.

Hence, the correct option is (c)

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