First slide

Pair of straight lines

Question

If ${ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0$ represents a pair of parallel lines then $\sqrt{(\frac{b}{a})}$

Moderate

A

$\sqrt{\frac{g^{2} - ac}{f^{2} - bc}}$
B

$\sqrt{\frac{f^{2} - bc}{g^{2} - ac}}$
C

$\frac{f^{2} - bc}{g^{2} - ac}$
D

$\frac{g^{2} - ac}{f^{2} - bc}$

Solution

By observing the options, consider the distance between the parallel lines
The distance between the parallel lines ${ax}^{2} + 2 hxy + {by}^{2} + 2 gx + 2 fy + c = 0$ is $2 \sqrt{\frac{g^{2} - ac}{a (a + b)}}$ or $2 \sqrt{\frac{f^{2} - bc}{b (a + b)}}$
Equate these two distances
$\begin{array}{l} 2 \sqrt{\frac{g^{2} - ac}{a (a + b)}} = 2 \sqrt{\frac{f^{2} - bc}{b (a + b)}} \\ \frac{g^{2} - ac}{a (a + b)} = \frac{f^{2} - bc}{b (a + b)} • squaring on both sides \\ \frac{g^{2} - ac}{a} = \frac{f^{2} - bc}{b} • Cancel out the common factors \\ \frac{b}{a} = \frac{f^{2} - bc}{g^{2} - ac} \end{array}$
Therefore, $\sqrt{\frac{b}{a}} = \sqrt{\frac{f^{2} - bc}{g^{2} - ac}}$

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