Pair Of Straight Lines Question 116
Question: The square of distance between the point of intersection of the lines represented by the equation $ ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 $ and origin, is
Options:
A) $ \frac{c(a+b)-f^{2}-g^{2}}{ab-h^{2}} $
B) $ \frac{c(a-b)+f^{2}+g^{2}}{\sqrt{ab-h^{2}}} $
C) $ \frac{c(a+b)-f^{2}-g^{2}}{ab+h^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let the lines represented by given equation be $ y=m_1x+c_1 $ and $ y=m_2x+c_2 $ .
Then $ ax^{2}+2hxy+by^{2}+2gx+2fy+c=0 $
$ =b(y-m_1x-c_1)(y-m_2x-c_2)=0 $
Comparing the coefficients of $ x^{2},\ xy,\ x,\ y $ and constant term, we get $ m_1m_2=\frac{a}{b},\ m_1+m_2=\frac{-2h}{b},\ m_1c_2+m_2c_1=\frac{2g}{b}, $
$ c_1+c_2=-\frac{2f}{b} $ and $ c_1c_2=\frac{c}{b} $
Also the point of intersection of $ y=m_1x+c_1 $ and $ y=m_2x+c_2 $ is $ ( \frac{c_2-c_1}{m_1-m_2},\frac{m_1c_2-m_2c_1}{m_1-m_2} ) $
Therefore, the square of distance of this point from origin is $ {{( \frac{c_2-c_1}{m_1-m_2} )}^{2}}+\frac{{{(m_1c_2-m_2c_1)}^{2}}}{{{(m_1-m_2)}^{2}}} $
$ =\frac{[{{(c_1+c_2)}^{2}}-4c_1c_2]+[{{(m_1c_2+m_2c_1)}^{2}}-4m_1m_2c_1c_2]}{{{(m_1+m_2)}^{2}}-4m_1m_2} $
Now putting the value defined above, we get the required distance i.e., $ \frac{-c(a+b)+f^{2}+g^{2}}{h^{2}-ab} $ .
BETA
