Straight Line Question 132
Question: Line L has intercepts a and b on the co-ordinate axes. When the axes are rotated through a given angle keeping the origin fixed, the same line L has intercepts p and q, then [IIT 1990; Kurukshetra CEE 1998]
Options:
A) $ a^{2}+b^{2}=p^{2}+q^{2} $
B) $ \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{p^{2}}+\frac{1}{q^{2}} $
C) $ a^{2}+p^{2}=b^{2}+q^{2} $
D) $ \frac{1}{a^{2}}+\frac{1}{p^{2}}=\frac{1}{b^{2}}+\frac{1}{q^{2}} $
Correct Answer: BShow Answer
Answer:
Solution:
Þ $ x( \frac{\cos \alpha }{a}+\frac{\sin \alpha }{b} )+y( \frac{\cos \alpha }{b}-\frac{\sin \alpha }{a} )=1 $…..(i) The intercepts made by (i) on the co-ordinate axes are given as p and q. Therefore $ \frac{1}{p}=\frac{\cos \alpha }{a}+\frac{\sin \alpha }{b} $ and $ \frac{1}{q}=\frac{\cos \alpha }{b}-\frac{\sin \alpha }{a} $ Squaring and adding, we get $ \frac{1}{p^{2}}+\frac{1}{q^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}} $ . Note: Students should remember this question as a formula.
BETA
