The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then $$1a2$$−$$1p2+1b2$$−$$1q2$$ is equal to

Question

The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a  fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then $$1a2$$−$$1p2+1b2$$−$$1q2$$ is equal to

Infinity Learn · Accepted Answer

Equation of the line L in the two coordinate systems is $$xa+yb=1$$,$$Xp+Yq=1where$$  (X$$, $$Y) are the new coordinates of a point (x$$, $$y) when the axes are rotated through a fixed angle, keeping the origin  fixed. As the length of the perpendicular from the origin has not $$changed.11a2+1b2=11p2+1q2$$⇒$$1a2+1b2=1p2+1q2or$$ $$1a2$$−$$1p2+1b2$$−$$1q2=0$$

The line $L$ has intercepts $a$ and $b$ on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If $p$ and $q$ are the intercepts of the line $L$ on the new axes, then $\frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}$ is equal to

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The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then 1a2−1p2+1b2−1q2 is equal to

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