Question

A line has intercepts a and b on the coordinate axes. When the axes are rotated through an angle $ \alpha $ , keeping the origin fixed, the line makes equal intercepts on the coordinate axes, then $ \tan \alpha = $
A. $ \dfrac{{a + b}}{{a - b}} $
B. $ \dfrac{{a - b}}{{a + b}} $
C. $ {a^2} - {b^2} $
D.None of these

Answer 1

Hint: Any point which lies on the x-axis is represented by (x,0) and that on the y-axis is represented by (0,y). When the axes are rotated, the coordinates of the points change. Use the formula for finding the new coordinates of the point and apply the given condition to find out the correct answer.

Complete step-by-step answer:
The line has intercepts a and b on the coordinate axes, the point at which this line cuts x-axis is (a,0) and the point at which the line cuts the y-axis is (0,b)
Now the axes are rotated by an angle $ \alpha $ .
The new coordinates of the point on the axis after it is rotated by an angle θ is given by, $ {x^1} = x\cos \theta + y\sin \theta $
And that of the y-axis is given by,
 $ {y^1} = y\cos \theta - x\sin \theta $
So the new coordinates of point A will be $ {A^1}(a\cos \alpha , - a\sin \alpha ) $ and that of point B will be $ {B^1}(b\sin \alpha ,b\cos \alpha ) $ .
The line joining these two points is given by the equation -
 $
\Rightarrow y - {y_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}(x - {x_1}) \\
\Rightarrow y + a\sin \alpha = \dfrac{{b\cos \alpha + a\sin \alpha }}{{b\sin \alpha - a\cos \alpha }}(x - a\cos \alpha ) \;
  $
Now, at the point of x-intercept, $ y = 0 $
So,
 $
\Rightarrow a\sin \alpha = \dfrac{{b\cos \alpha + a\sin \alpha }}{{b\sin \alpha - a\cos \alpha }}(x - a\cos \alpha ) \\
  x - a\cos \alpha = a\sin \alpha (\dfrac{{b\sin \alpha - a\cos \alpha }}{{b\cos \alpha + a\sin \alpha }}) \\
\Rightarrow x = \dfrac{{a\sin \alpha (b\sin \alpha - a\cos \alpha )}}{{b\cos \alpha + a\sin \alpha }} + a\cos \alpha \\
\Rightarrow x = \dfrac{{ab{{\sin }^2}\alpha - {a^2}\sin \alpha \cos \alpha + ab{{\cos }^2}\alpha + {a^2}\sin \alpha \cos \alpha }}{{b\cos \alpha + a\sin \alpha }} \\
\Rightarrow x = \dfrac{{ab({{\sin }^2}\alpha + {{\cos }^2}\alpha )}}{{b\cos \alpha + a\sin \alpha }} = \dfrac{{ab}}{{b\cos \alpha + a\sin \alpha }} \\
  $
Similarly for y-intercept, $ x = 0 $
So,
 $
\Rightarrow y = \dfrac{{(b\cos \alpha + a\sin \alpha )( - a\cos \alpha )}}{{b\sin \alpha - a\cos \alpha }} - a\sin \alpha \\
\Rightarrow y = \dfrac{{ - ab{{\cos }^2}\alpha - {a^2}\sin \alpha \cos \alpha - ab{{\sin }^2}\alpha + {a^2}\sin \alpha \cos \alpha }}{{b\sin \alpha - a\cos \alpha }} \\
\Rightarrow y = \dfrac{{ - ab({{\sin }^2}\alpha + {{\cos }^2}\alpha )}}{{b\sin \alpha - a\cos \alpha }} = \dfrac{{ - ab}}{{b\sin \alpha - a\cos \alpha }} \\
  $
Now, it is given in the question that the line makes equal intercepts on the axes,
Therefore, x-intercept = y-intercept
 $
\Rightarrow \dfrac{{ab}}{{b\cos \alpha + a\sin \alpha }} = \dfrac{{ - ab}}{{b\sin \alpha - a\cos \alpha }} \\
\Rightarrow \dfrac{1}{{b\cos \alpha + a\sin \alpha }} = \dfrac{{ - 1}}{{b\sin \alpha - a\cos \alpha }} \\
\Rightarrow b\sin \alpha - a\cos \alpha = - a\sin \alpha - b\cos \alpha \\
\Rightarrow a\sin \alpha + b\sin \alpha = a\cos \alpha - b\cos \alpha \\
\Rightarrow (a + b)\sin \alpha = (a - b)\cos \alpha \\
\Rightarrow \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{{a - b}}{{a + b}} \\
\Rightarrow \tan \alpha = \dfrac{{a - b}}{{a + b}} \\
  $
So, the correct answer is $ \dfrac{{a - b}}{{a + b}} $.

Note: A line passing through two points has an equation - $ y - {y_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}(x - {x_1}) $ .
When a line makes equal intercepts on the axes, it means that it cuts the x-axis and y-axis at an equal distance from the origin.