Question

If lines $a_{1}x+a_{2}y+a_3=0$ and $a_{4}x+a_{4}y+a_5=0$ are parallel, then $\frac{a_3}{a_5}$ may be equal to

• A. $\frac{a_2^2}{a_1{a}_4}$
• B. $\frac{a_2}{a_4}$
• C. $\frac{a_1}{a_4}$
• D. $\frac{a_4}{a_1{a}_2}$

Answer 1

The correct option is D $\frac{a_4}{a_1{a}_2}$

Given : $a_{1}x+a_{2}y+a_3=0$ and $a_{4}x+a_{4}y+a_5=0$ are parallel lines For the lines to be parallel,

$\frac{a_1}{a_4}=\frac{a_2}{a_4}\ne \frac{a_3}{a_5}$
Clearly,

$\frac{a_3}{a_5}\ne \frac{a_2}{a_4}.......\left(i\right)$

$\text{and}\frac{a_3}{a_5}\ne \frac{a_1}{a_4}$

$\text{Also,}\frac{a_1}{a_4}=\frac{a_2}{a_4}$

$\Rightarrow a_1=a_2$

Now, multiplying and dividing RHS of (i) by $a_1$, we get

$\frac{a_3}{a_5}\ne \frac{a_2\times a_1}{a_4\times a_1}$

$\Rightarrow \frac{a_3}{a_5}\ne \frac{a_2^2}{a_4{a}_1}$

$\frac{a_3}{a_5}\text{cannot be equal to}\frac{a_1}{a_4},\frac{a_2}{a_4}\text{and}\frac{a_2^2}{a_1{a}_4}$

$\text{Therefore,}\frac{a_3}{a_5}\text{may be equal to}\frac{a_4}{a_1{a}_4}$
Hence, the correct answer is option(d).